Chem 180 Set 1

Imagine that your water heater has broken, but you want to take a bath. You fill your bathtub with 25 kg of room-temperature water (about 25 °C). You figure that you can boil water on the stove and pour it into the bath to raise the temperature.
How much boiling water would you need to raise the bath to body temperature (about 37 °C)? Assume that no heat is transferred to the surrounding environment.

q=mSH ΔT (Where q is the amount of heat, m is the mass, SH is the specific heat, and ΔT is the change in temperature). q = (25000g)(4.184 J/g°C)(37°C-25°C) Note that the specific heat is a value obtained from an outside reference and the mass had to be converted to grams to cancel units with the specific heat. q = 1255200 J = 1255.2 KJ That gives you the amount of heat energy that you will need to raise your water to the desired temperature. Now, this value can be substituted in the above equation, along with the specific heat (constant) and a new delta T (equal to the difference in temperature between boiling water and the desired temperature of 37°C) and solve it for the mass of water necessary to loose that much heat. Note that the heat energy is negative because it will be the heat that is given up by the water instead of absorbed like above. q=mSH ΔT m = q /(SH ΔT) m = 1255200 J / [(4.184 J/g°C)(100°C-37°C)] m = 1255200 J / (263.592 J/g) m = 4762 g = 4.762 kg

The amount of boiling water required to raise the temperature of 25.0 kg of water in the bath to body temperature is 4.80 kg.
In this process, the heat lost by the boiling water is equal to the heat gained by the room-temperature water. How much heat was transferred in this process?

Using the equation q=mSH ΔT , where q is the heat, m is mass, SH is the specific heat capacity and ΔT is temperature change, and considering room temperature to be 25°C and body temperature to be 37°C, the heat transfer is as follows. q = 25 kg∙4.18J/g°C ∙ (37°C-25°C) q= 1254 kJ

A calorimeter is an insulated device in which a chemical reaction is contained. By measuring the temperature change, ΔT, we can calculate the heat released or absorbed during the reaction using the following equation:
q=specific heat×mass×ΔT

Or, if the calorimeter has a predetermined heat capacity, C, the equation becomes q=C×ΔT At constant pressure, the enthalpy change for the reaction, ΔH, is equal to the heat, qp; that is, ΔH=qp but it is usually expressed per mole of reactant and with a sign opposite to that of q for the surroundings. The total internal energy change, ΔE (sometimes referred to as ΔU), is the sum of heat, q, and work done, w: ΔE=q+w However, at constant volume (as with a bomb calorimeter) w=0 and so ΔE=qv.

A calorimeter contains 20.0 mL of water at 14.5°C . When 2.20 g of X (a substance with a molar mass of 69.0 g/mol ) is added, it dissolves via the reaction
X(s)+H2O(l)→X(aq)
and the temperature of the solution increases to 30.0 °C .
Calculate the enthalpy change, ΔH, for this reaction per mole of X.
Assume that the specific heat of the resulting solution is equal to that of water [4.18 J/(g⋅°C)], that density of water is 1.00 g/mL, and that no heat is lost to the calorimeter itself, nor to the surroundings.

Heat released by X raises the temp of water and X from 14.5°C to 30.0 °C Mass of water = Volume * density = 20 mL*1 g/mL = 20 g mass of x =2.2 g Heat required = (m1+m2)*C*(Tf-Ti) = (20+2.2)*4.18*(30-14.5) =1438.338 J 2.2 g of X releases 1438.338 J so,1 mol or 69 g will release = 1438.338 J*69/2.2 = 45111.51 J = 45.1 kJ since it is exothermic, it is negative, so ΔH = -45.1 kJ/mol.

Consider the reaction
C₁₂H₂₂O₁₁(s)+12O₂(g)→12CO₂(g)+11H₂O(l)
in which 10.0 g of sucrose, C₁₂H₂₂O₁₁, was burned in a bomb calorimeter with a heat capacity of 7.50 kJ/°C. The temperature increase inside the calorimeter was found to be 22.0 °C. Calculate the change in internal energy, ΔE, for this reaction per mole of sucrose.

C₁₂H₂₂O₁₁(s)+12O₂(g)→12CO₂(g)+11H₂O(l)→12CO₂(g)+11H₂O(l) m(sucrose) = 10g Molar mass(sucrose) = 342.296 g/mol moles of sucrose = 10g ∙ (1 mol/342.296 g/mol) = 0.0292 mol q(rxn)calorimeter =-CΔT =-7.50kJ/°C∙22°C =-165kJ ΔE=(-165kJ/10g sucrose) ∙ (342.296 g/1 mol sucrose) = -5648 kJ/mol

In the following experiment, a coffee-cup calorimeter containing 100 mL of H2O is used. The initial temperature of the calorimeter is 23.0 ∘C. If 6.30 g of CaCl2 is added to the calorimeter, what will be the final temperature of the solution in the calorimeter? The heat of solution ΔHsoln of CaCl2 is −82.8 kJ/mol.

since of heat of solution is negative, energy is released in this dissolving process. ΔHsoln=-q(calorimeter)-q(soln) q(calorimeter) = heat gained by calormeter q(soln) = heat gained by solution Assuming heat is not lost to the calorimeter, q(cal) = 0. Then, ΔHsoln=-q(soln) q(soln)=mSHΔT Assuming density of water to be 1 g/mL, the mass of water used = 100 g. Then mass of solution = 100 + 6 = 106 g. Assuming specific heat of solution to be same as that of water, C = 4.18 J/g°C Hence, q(soln)=(106 g).(4.18J/g°C)ΔT ΔHsoln=-82.8kJ/mol For 6 g of CaCl2 (molar mass = 110.98 g/mol), heat of solution is 6 gCaCl₂(1 mol CaCl₂/110.98g CaCl₂)(-82.8kJ/1 mol CaCl₂)=-4.48kJ since ΔH(soln)=-q(soln) -4.48kJ/mol=-(106 g)(4.18J/g°C)ΔT solving above equation, we get: ΔT = 10.11°C ΔT = T(f)-T(i) initial temperature = T(i) = 23° C Therefore, final temperature = T(f) = 23 + 10.11 = 33.11 °C

A volume of 90.0 mL of H2O is initially at room temperature (22.00 °C). A chilled steel rod at 2.00 ∘C is placed in the water. If the final temperature of the system is 21.00 °C , what is the mass of the steel bar?
Use the following values:
specific heat of water = 4.18 J/(g°C)
specific heat of steel = 0.452 J/(g°C)

q(H₂O) = mSHΔT =90 mL (4.18 J/g°C) -1 =-376.2 J q(SR) is equal to q(H₂O), so q(SR) = mSHΔT -376.2 J = m (0.452 J/g°C) ∙ -19°C -376.2 J = m (-8.588 J/g) m = 43.8 g

The specific heat of water is 4.18 J/(g⋅∘C). Calculate the molar heat capacity of water.

Molar mass of water(H2O) = 2 ∙ molar mass of (H atom) + 1 ∙ molar mass of (O atom) = 18.01528 g/mol Molar heat capacity of water = molar mass ∙ specific heat capacity of water = 18.01528 g/mol ∙ 4.18 J/(g∙°C) = 75.3 J/mol∙°C

A total of 2.00 mol of a compound is allowed to react with water in a foam coffee cup and the reaction produces 188 g of solution. The reaction caused the temperature of the solution to rise from 21.00 to 24.70 ∘C. What is the enthalpy of this reaction? Assume that no heat is lost to the surroundings or to the coffee cup itself and that the specific heat of the solution is the same as that of pure water.

q(sol) = m(sol) ∙ SH(sol) ∙ ΔT(sol) = 188g ∙ 4.18J/g°C ∙ 3.7 °C = 2907.6 J =2.9 kJ q(sol) = q(rxn) ΔH = -q(rxn)/moles = -2.9 kJ/2 mol = -1.45 kJ/mol

It takes 45.0 J to raise the temperature of an 9.00 g piece of unknown metal from 13.0∘C to 24.2 ∘C. What is the specific heat for the metal?

q=mSHΔT 45=9∙SH∙11.2 45=100.8∙SH SH=.446 J/g∙°C

The molar heat capacity of silver is 25.35 J/mol⋅∘C. How much energy would it take to raise the temperature of 9.00 g of silver by 12.1 ∘C?

q=mSHΔT 9 g Ag(1 mol Ag/107.87 g Ag) = .083 mol Ag q=(.083)(25.35 J/mol∙°C)(12.1 °C) = 25.6 J

When 50.0 mL of 1.00 M HCl is mixed with 50.0 mL of 1.00 M NaOH in a constant pressure calorimeter, a neutralization reaction takes place in which the temperature of the mixture increases from 25.0°C to 31.6°C. If the density of the mixture is assumed to be 1.00 g/mL and its specific heat is assumed to be 4.18 J/g.°C, what is ΔH</> in kJ for this reaction, assuming the calorimeter absorbs very little heat?

−55.2 kJ

What is the specific heat of lead if it takes 97.8 J to raise the temperature of a 75.1 g block by 10.5 ∘C?

q=mSHΔT 97.8 J = 75.1 g∙SH∙10.5°C 97.8 J = 788.55 ∙SH SH = .124 J/g∙°C

Assuming that Coca-Cola has the same specific heat as water [4.18 J/(g⋅∘C)], calculate the amount of heat in kilojoules transferred when one can (about 350 g) is cooled from 21 ∘C to 1 ∘C.

q=mSHΔT =350∙4.18∙-20 =-29260 J =-29.26 kJ

When 25.0 mL of 1.0 M H2SO4 is added to 50.0 mL of 1.0 M NaOH at 25.0 ∘C in a calorimeter, the temperature of the aqueous solution increases to 33.9 ∘C.
Assuming that the specific heat of the solution is 4.18 J/(g⋅∘C), that its density is 1.00 g/mL, and that the calorimeter itself absorbs a negligible amount of heat, calculate ΔH in kilojoules for the reaction.
H2SO4(aq)+2NaOH→2H2O(l)+Na2SO4(aq)

m = 25mL + 50 mL = 75 mL q(sol)=mSHΔT = 75 mL∙4.18J/g°C∙8.9°C = 2790.15 J =2.79 kJ q(sol)=q(rxn) 25 mL of 1.0 M H2SO4 is added to 50.0 mL of 1.0 M NaOH. 1 mol of H2SO4 reacts with 2 moles of NaOH, so only 25 mL of NaOH is consumed. E=0.025 mol∙ΔH ΔH = -q(rxn)/0.025 mol =2.79 kJ/0.025 mol =-111.6 kJ

What is the difference between heat capacity and specific heat?
Check all that apply.

Specific heat is the amount of heat necessary to raise the temperature of exactly 1 mole of a substance by exactly 1∘C.

Specific heat is the amount of heat necessary to raise the temperature of exactly 1 g of a substance by exactly 1∘C.

Heat capacity is the amount of heat necessary to raise the temperature of exactly 1 g of a substance by exactly 1∘C.

Heat capacity is the amount of heat necessary to raise the temperature of exactly 1 mole of a substance by exactly 1∘C.

Specific heat is the amount of heat required to raise the temperature of a substance a given amount.

Heat capacity is the amount of heat required to raise the temperature of a substance a given amount.

Specific heat is the amount of heat necessary to raise the temperature of exactly 1 g of a substance by exactly 1∘C. Heat capacity is the amount of heat required to raise the temperature of a substance a given amount.

Sodium metal is sometimes used as a cooling agent in heat-exchange units because of its relatively high molar heat capacity of 28.2 J/(mol⋅∘C).
What is the specific heat of sodium in J/(g⋅∘C)?

Cm=28.2 J/mol∙°C q=Cm∙moles of substance ∙ ΔT q=Cm ∙ (mass of sub/ molar mass of sub) ∙ ΔT SH ∙ m(Na) ∙ ΔT = Cm ∙ (m(Na)/mol m (Na)) ∙ ΔT∙ SH= Cm ∙ (1/23 g/mol) = 28.1/23 J/g°C SH = 1.23 J/g°C

When 1.045 g of CaO is added to 50.0 mL of water at 25.0 ∘C in a calorimeter, the temperature of the water increases to 32.3 ∘C.
Assuming that the specific heat of the solution is 4.18 J/(g⋅∘C) and that the calorimeter itself absorbs a negligible amount of heat, calculate ΔH in kilojoules for the reaction
CaO(s)+H2O(l)→Ca(OH)2(aq)

Density of H₂O at 25°C is 0.997 g/mL 50 mL ∙ 0.997 g H₂O/m1 = 49.85 g H₂O q= 49.85 g H₂O ∙ 4.18J/g°C ∙ 7.3°C =1562.8 J mw CaO = 56.077 g/mol mol CaO = 1.045/56.077 g/mol = 0.0186 mol 1562.85/0.0186 mol = 84000 J/mol = 84 kJ/mol Because it is exothermic, it would be -84 kJ/mol

When 0.187 g of benzene, C6H6, is burned in a bomb calorimeter, the surrounding water bath rises in temperature by 7.48 ∘C.
Assuming that the bath contains 250.0 g of water and that the calorimeter itself absorbs a negligible amount of heat, calculate combustion energies (ΔE) for benzene in both kilojoules per gram and kilojoules per mole.

m(benzine) = 0.187 g m(H₂O) = 250 g ΔT = 7.48°C SH(H₂O) = 4.18 J/g∙°C q=m(H₂O)∙SH(H₂O)∙ΔT(H₂O) =250 g ∙ 4.18 J/g∙°C ∙7.48°C =7816.6 J =7.82 kJ Heat released by benzine is – 7.82 kJ 0.187 g benzine = -7.82 kJ heat SO, 1 g benzine releases -7.82 kJ/0.187 g = -41.8 kJ/g Moles of C6H6 = 0.187 g ∙ (1 mol/78.108 g) =0.00239 mol C6H6 Heat released = -7.82 kJ / 0.00239 mol =-3272 kJ/mol

How much heat is evolved or absorbed (in kJ) if 10.0 g of Al(s) are used in the reaction 2Al(s) + 3O2(g) →2Al2O3(s) , which has a change in enthalpy of −3352 kJ?

According to the equation, 3352 kJ of heat is evolved for 2 moles of Al i.e 3352 kJ of heat is evolved for 2*27 g of Al i.e 3352 kJ of heat is evolved for 54 g of Al Thus, for 10 g of Al, using unitary method , heat evolved = (10/54)*3352 kJ = 620.74 kJ released

Calculate the standard enthalpy of the reaction, ΔH∘rxn, for the thermite reaction:
2Al(s)+Fe2O3(s)→2Fe(s)+Al2O3(s)
Elements in their standard state have an enthalpy of formation value of zero. The standard enthalpies of formation of Fe2O3 and Al2O3 are
ΔH∘f of Fe2O3(s)=−825.5 kJ/molΔH∘f of Al2O3(s)=−1675 kJ/mol

Calculate the enthalpy change for the thermite reaction:
2Al(s)+Fe2O3(s)→2Fe(s)+Al2O3(s), ΔH∘rxn=−850 kJ
when 12.0 mol of Al undergoes the reaction with a stoichiometrically equivalent amount of Fe2O3.

2 mol Al react with 1 mol Fe2O3 ΔH°(rxn)=-850 kJ ∙ (12 mol Al/2 mol Al) = -5100 kJ

What is the sign of ΔH for an exothermic reaction?

–

For an endothermic reaction?

+

Under what circumstances are ΔE and ΔH essentially equal?

When ΔV=0 (there are the same number of reactant and product gas molecules).

How much heat (in kilojoules) is evolved or absorbed in the reaction of 2.25 g of Na with H2O?
2Na(s)+2H2O(l)→2NaOH(aq)+H2(g) ΔH∘ = -368.4kJ.

2Na(s)+2H2O(l)———————–> 2NaOH(aq)+H2(g 2 x 23g 46 g Na gives ———————————> -368.4 kJ heat 1.35 g Na gives—————————–> x kJ heat x = 2.25 x -368.4 / 46 x = -18.02kJ The reaction is exotermic.

How much heat (in kilojoules) is evolved or absorbed in the reaction of 3.05 g of Fe2O3 with enough carbon monoxide to produce iron metal?
Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g) ΔH∘ = -24.8kJ.

The process would be considered as exothermic, since heat is being released. 3.05gm/159.6887 g/mol=0.473

What is the change in enthalpy for the reaction 2S(s) + 3O2(g) →2SO3(g), if given the following information: S(s) + O2(g) →SO2(g), change in enthalpy = −297 kJ; and 2SO3(g) →2SO2(g) + O2(g), change in enthalpy = +198 kJ?

Procedure for Using Hess’s Law

To understand how to use Hess’s law to find the enthalpy of an overall reaction. The change in enthalpy, ΔH, is the heat absorbed or produced during any reaction at constant pressure. Hess’s law states that ΔH for an overall reaction is the sum of the ΔH values for the individual reactions. For example, if we wanted to know the enthalpy change for the reaction 3Mn+3O2→3MnO2 we could calculate it using the enthalpy values for the following individual steps: Step 1: 4Al+3O2→2Al2O3 Step 2: 3Mn+2Al2O3→3MnO2+4Al Overall: 3Mn+3O2→3MnO2 If the enthalpy change is −3352 kJ/mol for step 1 and 1792 kJ/mol for step 2, then the enthalpy change for the overall reaction is calculated as follows: ΔH=−3352+1792=−1560 kJ/mol It is also important to note that the change in enthalpy is a state function, meaning it is independent of path. In other words, the sum of the ΔH values for any set of reactions that produce the desired product from the starting materials gives the same overall ΔH.

1. Now consider the following set of reactions:
2NO2→2NO + O2 ,ΔH=109 kJ/mol
N2O4→2NO + O2 ,ΔH=172 kJ/mol
The equations given in the problem introduction can be added together to give the following reaction:
overall: N2O4→2NO2
However, one of them must be reversed. Which one?

2. What is the enthalpy for reaction 1 reversed?
reaction 1 reversed: 2NO + O2→2NO2

3. What is the enthalpy for the following reaction?
overall: N2O4→2NO2

1. reaction 1: 2NO2→2NO + O2 because 2NO2 needs to be the product. 2. ΔH = -109kJ/mol because if you reverse the equation, you must reverse the sign of enthalpy 3. ΔH = 63 kJ/mol See picture

Calculate the enthalpy of the reaction
2NO(g)+O2(g)→2NO2(g)
given the following reactions and enthalpies of formation:
12N2(g)+O2(g)→NO2(g), ΔH∘A=33.2 kJ
12N2(g)+12O2(g)→NO(g), ΔH∘B=90.2 kJ

1/2N2(g) + O2(g) → NO2(g), ΔH∘A=33.2 kJ multiply this equation with 2 N2 + 2O2 → 2 NO2 , ΔH∘ = 66.4 kJ ↔ (1) 1/2N2(g) +1/2 O2(g)→ NO(g), ΔH∘B=90.2 kJ reverse this equation and multiply with 2 2NO → N2 + O2 , ΔH∘ = – 180.4 kJ ↔ (2) now add (1) + (2) N2 + 2O2→ 2 NO2 , ΔH∘ = 66.4 kJ ↔(1) 2NO → N2 + O2 , ΔH∘ = – 180.4 kJ ↔( 2) 2NO + O2 → 2 NO2 , ΔH∘ = -114 kJ

For which of the following reactions is ΔH∘rxn equal to ΔH∘f of the product(s)?
You do not need to look up any values to answer this question.
C(s,graphite)+O2(g)→CO2(g)
CaCO3(g)→CaO+CO2(g)
Li(s)+12F2(l)→LiF(s)
2Li(s)+F2(g)→2LiF(s)
CO(g)+12O2(g)→CO2(g)
Li(s)+12F2(g)→LiF(s)

C(s,graphite)+O2(g)→CO2(g) Li(s)+12F2(g)→LiF(s) For both of these reactions, the calculation of ΔH∘rxn would look something like ΔH∘rxn=[1×ΔH∘f(product)]−[0+0] If the coefficient for the product were not 1, or if the reactants were not standard-state elements, then the enthalpy change of the reaction would not be equal to ΔH∘f for the product.

What is the enthalpy change for the reaction MgO(s) + 2HCl(g) →MgCl2(s) + H2O(l)? Enthalpies of formation are −601.7 kJ/mole for MgO, −92.3 kJ/mole for HCl, −641.6 kJ/mole for MgCl2, and −285.8 kJ/mole for H2O(l).

−141.1 kJ The enthalpy change for a reaction = sum of enthalpy of products – sum of enthalpy of reactants thus, Enthalpy change in this case = (-641.6 – 285.8) – (-601.7 + 2 x -92.3) = -141.1 kJ

Calculate the enthalpy of the reaction
4B(s)+3O2(g)→2B2O3(s)
given the following pertinent information:
1 B2O3(s)+3H2O(g)→3O2(g)+B2H6(g), ΔH∘A=+2035 kJ
2 2B(s)+3H2(g)→B2H6(g), ΔH∘B=+36 kJ
3 H2(g)+12O2(g)→H2O(l), ΔH∘C=−285 kJ
4 H2O(l)→H2O(g), ΔH∘D=+44 kJ

ΔH∘ = -2552 kJ 1 flip and multiply by 2 2 multiply by 2 3 flip and multiply by 6 4 flip and multiply by 6

The combustion of pentane, C5H12, occurs via the reaction
C5H12(g)+8O2(g)→5CO2(g)+6H2O(g)
with heat of formation values given by the following table:
Substance ΔH∘f (kJ/mol)
C5H12 (g) -119.9
CO2(g) −393.5
H2O(g) −241.8
Calculate the enthalpy for the combustion of 1 mole of pentane.

ΔH∘ =products-reactants =[5(-393.5)+6(−241.8)]-(-119.9+0) =-3298.5 kJ

How is the standard state of an element defined?

It is the most stable form of an element at 1 atm and the specified temperature, usually 25 ∘C.

What is Hess’s law?
Why does it "work"?

The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction. Because of the law of conservation of energy.

Why do elements always have ΔH∘f=0?

The standard state of elements is the reference point for measuring of enthalpy changes.

One possible use for the cooking fat left over after making french fries is to burn it as fuel.
Formula = C51H88O6
Density = 0.94g/mL
ΔH∘f = – 1310kJ/mol.
Write a balanced equation of the combustion of cooking fat
Use the data above to calculate the amount of energy released (in kilojoules per milliliter) from the combustion of cooking fat:

C51H88O6(l)+70O2(g)→51CO2(g)+44H2O(l) ΔH∘f [C51H88O6(l)] = -1310 kJ/mol ΔH∘f [O2 (g)] = 0 kJ/mol ΔH∘f[CO2 (g)] = -393.5 kJ/mol ΔH∘f[H2O (l)] = -285.8 kJ/mol ΔH∘rxn = ΔH∘f(products) – ΔH∘f(reactants) = 51 x -393.5 kJ/mol + [ 44 x -285.8 kJ/mol] – [-1310 kJ/mol + 70 x 0 ] = -31333.7 kJ/mol ΔH∘rxn = -31333.7 kJ/mol Negative sign indicates , heat energy is released. Molar mass of C51H88O6 = 796 g density = 0.94 g/mL volume = mass/ density = 796 g / 0.94 g/mL = 846.8 mL Heat energy released per mL = 31333.7 kJ / 846.8 mL = 37.0 kJ/mL Heat energy released per mL = 37.0 kJ/mL

1) A+B→C ΔH∘ = -80 kJ
2) C+B→D ΔH∘ = -40 kJ
What is the equation for the net reaction?
What is the ΔH∘ for the net reaction?

A+2B→D ΔH° = -120 kJ

The diagram shows three energy levels. The energies of which substances are represented by each? Which arrow on the diagram corresponds to which step? Which arrow corresponds to the net reaction?

Sulfuric acid (H2SO4), the most widely produced chemical in the world, is made by a two-step oxidation of sulfur to sulfur trioxide, SO3, followed by reaction with water.
Calculate ΔH∘f for SO3 in kJ/mol, given the following data:
S(s)+O2(g)→SO2(g); ΔH∘ = -296.8 kJ
SO2(g)+1/2O2(g)→SO3(g); ΔH∘ = -98.9 kJ

S(s)+O2(g)→SO2(g); ΔH∘ = -296.8 kJ + SO2(g)+1/2O2(g)→SO3(g); ΔH∘ = -98.9 kJ ———————————————– S(s)+³/₂O2(g)→SO3(g); ΔH∘ = -395.7 kJ

Calculate ΔH∘f (in kilojoules per mole) for benzene, C6H6, from the following data:
2C6H6(l)+15O2(g)→12CO2(g)+6H2O(l) ΔH∘ = -6534kJ
ΔH∘f (CO2) = -393.5kJ/mol
ΔH∘f (H2O) = – 285.8kJ/mol

C(s)+CO2(g)→2CO(g)

2H2O2(aq)→2H2O(l)+O2(g)

Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)

Given the standard heats of formation shown in Appendix B in your textbook, what is ΔH∘ in kilojoules for the reaction
3N2O4(g)+2H2O(l)⟶4HNO3(aq)+2NO(g)?

C8H8(l)+10O2(g)→8CO2(g)+4H2O(l)

CalculateΔH∘f for styrene in kJ/mol.
ΔH∘f[CO2(g)]=−393.5kJ/mol;
ΔH∘f[H2O(l)]=−285.8kJ/mo

ΔH∘f = 104 kJ/mol

In chemical reactions, heat is converted into chemical energy (the potential energy stored in chemical bonds) or vice versa. Thus, enthalpy change for a reaction can be approximated from
ΔH∘=D(reactants)−D(products)
where D represents bond dissociation energies of the reactants and products, respectively.
The table below contains the bond dissociation energies for common bonds. The information can be used to calculate the energy needed to break all the bonds in a given molecule.

Due to the variation in literature values of thermodynamic quantities:
Assume that these values sufficiently represent the bonds in the compounds involved in the following questions
Be sure that you use the values given in the table (above) in the introduction for your calculations in this item.
Calculate the bond dissociation energy for the breaking of all the bonds in a mole of methane, CH4.

Use the data in the table given below to calculate an approximate ΔH∘ in kilojoules for the synthesis of hydrazine from ammonia:
2NH3(g)+Cl2(g)⟶N2H4(g)+2HCl(g)

-81 kJ

Use the bond dissociation energies in Table 8.3 in the textbook to calculate an approximate ΔH∘ (in kilojoules) for the reaction of ethylene with hydrogen to yield ethane.
H2C=CH2(g)+H2(g)→CH3CH3(g)

-6 kJ

Part A
Calculate an approximate heat of combustion for butane in kilojoules by using the bond dissociation energies in the table. (The strength of the O=O bond is 498 kJ/mol, and that of a C=O bond in CO2 is 804 kJ/mol.)

See next card for answer

Part B
Calculate an approximate heat of combustion for butane in kilojoules by using the bond dissociation energies in the table. (The strength of the O=O bond is 498 kJ/mol, and that of a C=O bond in CO2 is 804 kJ/mol.)

± Enthalpy, Entropy, and Spontaneity

The spontaneity of a reaction depends both on the enthalpy change, ΔH, and entropy change, ΔS. Reactions that release energy produce more stable products, and the universe tends toward disorder. Thus, an exothermic reaction with a positive entropy change will always be spontaneous. Mathematically, this relationship can be represented as ΔG=ΔH−TΔS where ΔG is the change in Gibbs free energy and T is the Kelvin temperature. If ΔG is negative, then the reaction is spontaneous. If ΔG is positive, then the reaction is nonspontaneous as written but spontaneous in the reverse direction.

What can be said about an endothermic reaction with a negative entropy change?

The reaction is spontaneous in the reverse direction at all temperatures.

What can be said about an exothermic reaction with a negative entropy change?

The reaction is spontaneous at low temperatures.

If ΔH = -70.0 kJ and ΔS = -0.200 kJ/K , the reaction is spontaneous below a certain temperature. Calculate that temperature.

Set ΔG=0 ΔG=ΔH-ΔST 0=-70-(-0.2T) 70=0.2T T=350 Reaction is spontaneous below 350K

Which of the following statements is true?

The following substances are listed in order of increasing entropy: helium, gasoline, and sugar.

The following substances are listed in order of increasing entropy: H2O(l), H2O(s), and H2O(g).

The following substances are listed in order of increasing entropy: concrete, cooking oil, and water vapor.

The following substances are listed in order of decreasing entropy: neon, iron, and mercury.

The following substances are listed in order of increasing entropy: concrete, cooking oil, and water vapor. solid < liquid < gas

Ethane, C2H6, can be prepared by the reaction of acetylene, C2H2, with hydrogen.
C2H2(g)+2H2(g)→C2H6(g)

Explain why ΔS∘ is negative.

Negative The reaction decreases the number of moles of gaseous molecules.

Which free energy change is correctly classified as either spontaneous or non-spontaneous?

0.0 kJ, spontaneous
+234 kJ, spontaneous
−2.3 × 10−3 kJ, nonspontaneous
+10.0 kJ, non-spontaneous

+10.0 kJ, nonspontaneous

Which of the following reactions are spontaneous under standard conditions at 25 ∘C, and which are nonspontaneous?

(a) AgNO3(aq)+NaCl(aq)⟶AgCl(s)+NaNO3(aq)ΔG∘=−55.7kJ

(b) 2C(s)+2H2(g)⟶C2H4(g)ΔG∘=68.1kJ

(a) is spontaneous, (b) is non-spontaneous

Is the Haber process for the industrial synthesis of ammonia spontaneous or nonspontaneous under standard conditions at 25 ∘C?
N2(g)+3H2(g)→2NH3(g) ΔH∘=−92.2 kJ; ΔS∘=−199 J/K

At what temperature (∘C) does the changeover occur?

spontaneous 190°C

What does entropy measure?

Entropy is a measure of molecular randomness.

What are the two terms that make up the free-energy change for a reaction, ΔG.

Which of the two is usually more important?

ΔH and ΔS. ΔH

How is it possible for a reaction to be spontaneous yet endothermic?

If the TΔS term is larger than ΔH.

Tell whether the entropy changes for the following processes are likely to be positive or negative:

(a) The fizzing of a newly opened can of soda

(b) The growth of a plant from seed

(a) is positive, (b) is negative

Tell whether the free-energy changes, ΔG, for the process below are likely to be positive, negative, or zero.

The conversion of liquid water to water vapor at 100 ∘C.

ΔG is to be equal to zero.

Tell whether the free-energy changes, ΔG, for the process below are likely to be positive, negative, or zero.

The freezing of liquid water to ice at 0 ∘C.

ΔG is to be equal to zero.

Tell whether the free-energy changes, ΔG, for the process below are likely to be positive, negative, or zero.

The eroding of a mountain by a glacier.

ΔG is to be negative.

One of the steps in the cracking of petroleum into gasoline involves the thermal breakdown of large hydrocarbon molecules into smaller ones. For example, the following reaction might occur:
C11H24⟶C4H10+C4H8+C3H6

Is ΔS for this reaction likely to be positive or negative?

Choose the correct explanation.

Positive The reaction increases the total number of molecules.

Tell whether reactions with the following values of ΔH and ΔS are spontaneous or nonspontaneous and whether they are exothermic or endothermic:

ΔH=−128 kJ, ΔS=35 J/K at 500 K

ΔH = -ve (exothermic reaction) ΔG = ΔH – T x ΔS ΔG = (-128 kJ) – (500 x 0.035) = -128 – 17.5 = – 145.5 kJ (spontaneous reaction)

ΔH=+67 kJ, ΔS=−140 J/K at 250 K

ΔH = +ve (endothermic reaction) ΔG = ΔH – T x ΔS ΔG = (+67 kJ) – (250 x – 0.14) = 67 + 35 = + 102 kJ (non-spontaneous reaction)

ΔH=+75 kJ, ΔS=95 J/K at 800 K

ΔH = +ve (endothermic reaction) ΔG = ΔH – T x ΔS ΔG = (+75 kJ) – (800 x – 0.095) = 75 + 76 = + 151 kJ spontaneous reaction)

Write balanced equation for the combustion reaction of propanol (C3H8O) with oxygen to give CO2 and H2O.

2C₃H₈O+9O₂→6CO₂+8H₂O

Write balanced equation for the combustion reaction of stearic acid (C18H36O2) with oxygen to give CO2 and H2O.

C₁₈H₃₆O₂+26O₂→18CO₂+18H₂O

Imagine a reaction that results in a change in both volume and temperature. (see picture)

Has any work been done?
If so, is its sign positive or negative?
Has there been an enthalpy change?
If so, what is the sign of ΔH?
Is the reaction exothermic or endothermic?

Volume of the reaction vessel increased means the work done by the reaction is negative Because formula to calculate the work is W=-P*deltaV Delta V= V2-V1 When the delta V is positive then work is negative The temperature of the reaction mixture increases means the reaction is exothermic therefore the sign for the delta H is negative W= -, Delta H = -, exothermic

Acetylene, C2H2, reacts with H2 in two steps to yield ethane, CH3CH3:
(1) HC≡CH+H2⟶H2C=CH2 ΔH∘=−175.1kJ
(2) H2C=CH2+H2⟶CH3CH3 ΔH∘=−136.3kJ
Net HC≡CH+2H2⟶CH3CH3 ΔH∘=−311.4kJ

Which arrow (a, b, c) in the Hess’s law diagram corresponds to which step, and which arrow corresponds to the net reaction? Where are the reactants located on the diagram, and where are the products located?

Draw a Hess’s law diagram for the reaction of ethyl alcohol (CH3CH2OH) with oxygen to yield acetic acid (CH3CO2H).
(1) CH3CH2OH(l)+1/2O2(g)→CH3CHO(g)+H2O(l) ΔH∘=−174.2kJ
(2) CH3CHO(g)+1/2O2(g)→CH3CO2H(l) ΔH∘=−318.4kJ
Net CH3CH2OH(l)+O2(g)→CH3CO2H(l)+H2O(l) ΔH∘=−492.6kJ

A reaction is carried out in a cylinder fitted with a movable piston. The starting volume is V=5.00 L, and the apparatus is held at constant temperature and pressure.

Does V increase, decrease, or remain the same after reaction? Assume that ΔH = -35.0 kJ and ΔE = -34.6 kJ.

Calculate the final volume V.

Use example in pic, but sub -34.6 for -34.4. V decreases. V = 1.04 L

The addition of H2 to C=C double bonds is an important reaction used in the preparation of margarine from vegetable oils. If 50 mL of H2 and 50 mL of ethylene (C2H4) are allowed to react at 1.2 atm , the product ethane (C2H6) has a volume of 50 mL .

Calculate the amount of PV work done.
C2H4(g)+H2(g)→C2H6(g)

Tell the direction of the energy flow.

Total initial volume, V1 = 50 mL + 50 mL =100 mL = 0.1 L Final volume, V2 = 50 mL = 0.05 L ΔV = 0.05 L – 0.1L = – 0.05 L W = – PΔV = – (1.2 atm)(-0.05 L) = 0.06 L atm = 0.06 L atm * 101.325 J/ 1L atm =6.0795 J Here work done has positive value that means workdone on the system by surroundings. So energy flows fromsurroundings to system.

What is the difference between the internal energy change ΔE and the enthalpy change ΔH? Which of the two is measured at constant pressure, and which at constant volume?

What is the enthalpy change (ΔH) for a reaction at a constant pressure of 1.00 atm if the internal energy change (ΔE) is 48.0 kJ and the volume increase is 16.0 L ? (1 L×atm=101.325J.)

The familiar "ether" used as an anesthetic agent is diethyl ether, C4H10O. Its heat of vaporization is +26.5 kJ/mol at its boiling point.

How much energy (in kilojoules) is required to convert 230 mL of diethyl ether at its boiling point from liquid to vapor if its density is 0.7138 g/mL?

Use example in pic, and sub 230 mL for 100 mL E = 58.7 kJ

Aluminum metal reacts with chlorine with a spectacular display of sparks:
2Al(s)+3Cl2(g)→2AlCl3(s) ΔH∘ = -1408.4 kJ.

How much heat (in kilojoules) is released on reaction of 5.45 g of Al?

Molar mass of Al is 27 g/mol According to the balanced equation , 2 moles of Al on reaction releases 1408.4 kJ of energy OR 2×27 g of Al on reaction releases 1408.4 kJ of energy 5.60 g of Al on reaction releases Z kJ of energy [1408.4/( 2∙ 27)]∙ 5.45=142.1 kJ So the heat released on reaction of 5.45g of Al is 142.1 kJ

How much heat in kilojoules is evolved or absorbed in the reaction of 269.0 g of calcium oxide with enough carbon to produce calcium carbide?

CaO(s)+3C(s)→CaC2(s)+CO(g) ΔH∘ = 464.6kJ

Is the process exothermic or endothermic?

moles = mass / molar mass mass of CaO = 269.0 g molar mass of CaO = 56.0774 g /mol moles of CaO = 269.0 / 56.0774 moles of CaO = 4.642857 CaO + 3C —> CaC2 + CO 1 mole of CaO —> 464.6 kJ let 4.642857 mole of CaO —> y kJ y = 4.642857 x 464.6 y = 2231.739 heat is positive , so it absorbed 2231.739 kJ of heat is absorbed endothermic

Given the standard heats of formation
Δ H∘f CaO=−634.9 kJ/mol
Δ H∘f CO2=−393.5 kJ/mol
Δ H∘f CaCO3=−1207.6 kJ/mol
What is Δ H∘ in kilojoules for the reaction CaCO3(s)⟶CaO(s)+CO2(g)?

CaCO3(s)⟶CaO(s)+CO2(g)? Hrxn = Hproducts – Hreactants Hrxn = (−634.9 + −393.5) – (−1207.6) = 179.2 kJ/mol

Tell whether the entropy changes, ΔS, for the following processes are likely to be positive or negative.

The conversion of liquid water to water vapor at 100 ∘C

The freezing of liquid water to ice at 0 ∘C

The eroding of a mountain by a glacier

In this process ΔS is likely to be positive. In this process ΔS is likely to be negative. In this process ΔS is likely to be positive.

When a bottle of perfume is opened, odorous molecules mix with air and slowly diffuse throughout the entire room.

Is ΔG for the diffusion process positive, negative, or zero?

What about ΔH for the diffusion?

What about ΔS for the diffusion?

ΔG for this process is negative. ΔH for this process is approximately equal to zero. ΔS for this process is positive.

Tell which reaction(s) with the following values of ΔH and ΔS are spontaneous.

ΔH = -48 kJ, ΔS = +135 J/K at 300 K
ΔH = -48 kJ, ΔS = -135 J/K at 300 K
ΔH = +48 kJ, ΔS = +135 J/K at 300 K
ΔH = +48 kJ, ΔS = -135 J/K at 300 K

ΔH = -48 kJ, ΔS = +135 J/K at 300 K ΔH = -48 kJ, ΔS = -135 J/K at 300 K ΔH = -, ΔS = + is always spontaneous ΔH = -, ΔS = – is spontaneous at low temperatures

Tell which reaction(s) with the following values of ΔH and ΔS are exothermic.

ΔH = -48 kJ, ΔS = +135 J/K at 300 K
ΔH = +48 kJ, ΔS = -135 J/K at 300 K
ΔH = +48 kJ, ΔS = +135 J/K at 300 K
ΔH = -48 kJ, ΔS = -135 J/K at 300 K

ΔH = -48 kJ, ΔS = +135 J/K at 300 K ΔH = -48 kJ, ΔS = -135 J/K at 300 K ΔH = – is exothermic

(a) ΔH = – 48 kJ; ΔS = + 135 J/K at 400 K
(b) ΔH = – 48 kJ; ΔS = – 135J/K at 400 K
(c) ΔH = + 48 kJ; ΔS = + 135 J/K at 400 K
(d) ΔH= + 48 kJ; ΔS = – 135 J/K at 400 K

Which of the reactions are spontaneous at all temperatures?

Which of the reactions are nonspontaneous at all temperatures?

Which of the reactions have an equilibrium temperature?

a d b, c

When 1.20 g of magnesium metal is allowed to react with 200 mL of 6.00 M aqueous HCl, the temperature rises from 25.0 ∘C to 43.7 ∘C.

Calculate ΔH in kilojoules for the reaction, assuming that the heat capacity of the calorimeter is 776 J/∘C, that the specific heat of the final solution is the same as that of water [(4.18 J/ (g⋅∘C)], and that the density of the solution is 1.00 g/mL.

Use example. -613 kJ/mol

Find Δ H∘ in kilojoules for the reaction of nitric oxide with oxygen, 2NO(g)+O2(g)⟶N2O4(g), given the following data:
N2O4(g)⟶2NO2(g)ΔH∘=55.3kJNO(g)+1/2O2(g)⟶NO2(g)ΔH∘=−58.1kJ

2NO(g)+O2(g)⟶N2O4(g) Given N2O4(g)⟶2NO2(g) ΔH∘=55.3kJ NO(g)+1/2O2(g)⟶NO2(g)ΔH∘=−58.1kJ invert(1) since we need N2O4 in the product side 2NO2(g)⟶N2O4(g) ΔH∘= -55.3kJ NO(g)+1/2O2(g)⟶NO2(g)ΔH∘=−58.1kJ Multiply (2) by 2x to calcal NO2 2NO2(g)⟶N2O4(g) ΔH∘= -55.3kJ 2NO(g)+O2(g)⟶2NO2(g)ΔH∘=2*(−58.1) = -116.2 kJ Add both equations 2NO2(g)+2NO(g)+O2(g)⟶N2O4(g) + 2NO2(g) Hrxn = -55.3 + -116.2 = – 171.5 kJ cnacel common terms 2NO(g)+O2(g)⟶N2O4(g) Hrxn = – 171.5 kJ

What is the melting point of benzene in kelvin if Δ Hfusion=9.95kJ/mol and Δ Sfusion=35.7J/ (K×mol)?

Hess’s law can be used to calculate reaction enthalpies for hypothetical processes that can’t be carried out in the laboratory. Set up a Hess’s law cycle that will let you calculate Δ H∘ for the conversion of methane to ethylene:
2CH4(g)⟶C2H4(g)+2H2(g)
You can use the following information:
2C2H6(g)+7O2(g)⟶4CO2(g)+6H2O(l)ΔH∘=−3120.8kJCH4(g)+2O2(g)⟶CO2(g)+2H2O(l)ΔH∘=−890.3kJC2H4(g)+H2(g)⟶C2H6(g)ΔH∘=−136.3kJH2O(l)ΔH∘f=−285.8kJ/mol

4 CO2 (g) + 6 H20 (l) –> 2C2H6 (g) + 7O2 (g) delta H = 3120.8 kJ 4CH4 (g) + 8O2 (g) –> 4CO2 (g) + 4H2O (l) delta H = -3561.2 kJ 2 C2H6 (g) –> 2 C2H4 (g) + 2H2 (g) delta H = 272.6 kJ H2 + 1/2O2 —> H2O delta H = -285.8 kJ flip the equation and multiply by 2. add all four equations to get: 4CH4 (g) –> 2C2H4 (g) + 4H2 divide by 2 2 CH4 (g) –> C2H4 (g) + 2H2

Set up a Hess’s law cycle, and use the following information to calculate ΔH∘f for aqueous nitric acid, HNO3(aq). You will need to use fractional coefficients for some equations.
3NO2(g)+H2O(l)→2HNO3(aq)+NO(g) ΔH∘ = -137.3 kJ
2NO(g)+O2(g)→2NO2(g) ΔH∘ = -116.2 kJ
4NH3(g)+5O2(g)→4NO(g)+6H2O(l) ΔH∘ = -1165.2 kJ
NH3(g) ΔH∘f = -46.1 kJ
H2O(l) ΔH∘f = -285.8 kJ

3NO2(g) + H2O(l) → 2HNO3(aq)+ NO(g) ΔH = – 137.3 kJ 2NO(g) +O2(g) →2NO2(g) ΔH = – 116.2 kJ 4NH3(g) + 5O2(g) → 4NO(g) +6H2O(l) ΔH = -1165.2 kJ 1/2N2(g) + 3/2H2(g) →NH3(g) ΔH = – 46.1 kJ H2(g) + 1/2O2(g) →H2O(l) ΔH = – 285.8 kJ ΔHf of N2(g),H2(g) and O2(g)is 0 ½∙ 3/2NO2(g) +1/2H2O(l) →HNO3(aq) + 1/2NO(g) ΔH = – 68.65 kJ ¾∙ 3/2NO(g) +3/4O2(g) →3/2NO2(g) ΔH = – 87.15 kJ ¼∙ NH3(g) +5/4O2(g) →NO(g) + 3/2H2O(l) ΔH = -291.30 kJ 1/2N2(g) + 3/2H2(g) → NH3(g) ΔH = – 46.1 kJ (reverse) H2O(l)→ H2(g) +1/2O2(g) ΔH = + 285.8 kJ equals 1/2N2(g) + 1/2H2(g) + 3/2O2(g)→ HNO3(aq) ΔHf = – 68.65 – 87.15 – 291.3 – 46.1 + 285.8 = -207.4 kJ

Determine the sign of ΔS° for each of the following:
I. The mixing of two gases at a given temperature and pressure
II. C(s) + 2 H2O(g) → CO2(g) + 2 H2(g)

ΔS° is positive for I and positive for II.

For an explosion in an open vessel, one would expect

ΔH to be negative and ΔE to be less than ΔH.

Determine the sign of ΔS° for each of the following:
I. C6H6(s) → C6H6(l)
II. 2 SO2(g) + O2(g) → 2 SO3(g)

∆S° should be positive for I and negative for II.

Calculate the total quantity of heat required to convert 25.0 g of liquid CCl4(l) from 25.0°C to gaseous CCl4 76.8°C (the normal boiling point for CCl4)? The specific heat of CCl4(l) is 0.857 J/(g ⋅ ºC) its heat of fusion is 3.27 kJ/mol and its heat of vaporization is 29.82 kJ/mol.

25.0g(0.857J/gºC)(76.8-25ºC)= 1109.8 or1110 J change to kJ: 1110J(1kJ/1000J)=1.11kJ Since your vaporization heat is in kJ/mol, you have to figure out how many moles you have. Molar mass of CCl4 is 12.00+4(35.453)=12.00+141.8= 153.8g/mol You have 25.0g(1mole/153.8g)= 0.1625 moles Heat to vaporize= 0.1625moles(29.82kJ/mole)= 4.85 kJ Add this to heat needed to raise to bp: 4.85kJ + 1.11kJ= 5.96kJ

Calculate the enthalpy of combustion per mole for C6H12O6. Assume that the combustion products are CO2(g) and H2O(l).

Species ΔH°f, kJ/mol
C6H12O6(s) -1260
O2(g) 0
CO2(g) -393.5
H2O(l) -285.8

The values of ΔH°f for the three states of benzene are approximately -22 kcal/mol, -11 kcal/mol, and 20 kcal/mol. Which is the value for solid benzene?

-22 kcal/mol

the given average bond dissociation energies, D, to estimate ΔH for the following reaction:

CH3I(g) + H2(g) →CH4(g) + HI(g)
Bond D, kJ/mol
C-I 240
H-I 298
H-H 436
C-H 410

-32

A process is carried out at constant pressure. Given that 0 &gt; ΔH &gt; ΔE,
A process is carried out at constant pressure. Given that 0 &gt; ΔH &gt; ΔE,
the system loses heat and contracts during the process.
the system absorbs heat and expands during the process.
the system absorbs heat and contracts during the process.
the system loses heat and expands during the process

the system loses heat and expands during the process.

If a reaction with a negative value of ΔS is nonspontaneous at constant temperature and pressure,
If a reaction with a negative value of ΔS is nonspontaneous at constant temperature and pressure,
ΔG is negative and ΔH may be positive or negative.
ΔG is positive and ΔH may be positive or negative.
ΔG is negative and ΔH is positive.
ΔG is positive and ΔH is positive.

ΔG is positive and ΔH may be positive or negative.

…

…

When 4.000 moles of H2(g) reacts with 2.000 mol of O2(g) to form 4.000 mol of H2O(l) at 25°C and a constant pressure of 1.00 atm. If 273.2 kJ of heat are released during this reaction, and PΔV is equal to -14.80 kJ, then

ΔH° = -273.2 kJ and ΔE° = -258.4 kJ.

The reaction represented in the above drawing is likely to be

The reaction represented in the above drawing is likely to be
nonspontaneous at all temperatures.
nonspontaneous at low temperatures and spontaneous at high temperatures.
spontaneous at low temperatures and non spontaneous at high temperatures.
spontaneous at all temperatures.

spontaneous at low temperatures and non spontaneous at high temperature

Given: 4 NO2(g) + O2(g) → 2 N2O5(g) H° = -110.2 kJ
find ΔH° for N2O5(g) → 2 NO2(g) + 1/2 O2(g).

55.1 kJ

Imagine a reaction that results in a change in both volume and temperature, as shown in the diagram below. What is the sign of the work being done and the sign of the enthalpy change involved in this reaction?

w = + and ΔH = +

Reactant R reacts with reactant S in two steps to yield product Z.

The intermediate, T, + S is represented by

line E.

Which equation represents the reaction whose ΔH, represents the standard enthalpy of formation of CHCl3(l) at 25°C? (i.e., for which is ΔH = ΔH°f of CHCl3)

C(s) + 1/2 H2(g) + 3/2 Cl2(g) → CHCl3(l)

For the conversion of water to ice at 25°C and 1 atm,
For the conversion of water to ice at 25°C and 1 atm,
ΔG is positive and ΔH is positive.
ΔG is negative and ΔH is positive.
ΔG is positive and ΔH is negative.
ΔG is negative and ΔH is negative

ΔG is negative and ΔH is negative.

When 1.00 mol of benzene is vaporized at a constant pressure of 1.00 atm and at its normal boiling point of 80.1°C, 33.9 kJ are absorbed and PΔV for the vaporization process is equal to 2.90 kJ, then

ΔE = 31.0 kJ and ΔH = 33.9 kJ.

Reactant R reacts with reactant S in two steps to yield product Z.

Step (1) in the reaction is represented by

arrow b

Most chemical reactions are carried out in one of two ways:
I. in an open vessel at constant atmospheric pressure
II. in a closed vessel
Which is true?

ΔH = q for condition I and ΔE = q for condition II

The following drawing is a representation of a reaction for which ΔH° = +62 kJ. This reaction is likely to be

nonspontaneous at low temperatures and spontaneous at high temperatures.