Problem 1

Hydrogen peroxide decomposes according to the following thermochemical reaction:

H2O2(l) → H2O(l) + 1/2 O2(g); ΔH = -98.2 kJ

Calculate the change in enthalpy, ΔH, when 1.00 g of hydrogen peroxide decomposes.

Solution

The thermochemical equation tells us that ΔH for the decomposition of 1 mole of H2O2 is -98.2 kJ, so

this relationship can be used as a conversion factor. Using the Periodic Table, the molecular mass of

H2O2 is 34.0, which means that 1 mol H2O2 = 34.0 g H2O2.

Using these values:

ΔH = 1.00 g H2O2 x 1 mol H2O2 / 34.0 g H2O2 x -98.2 kJ / 1 mol H2O2

ΔH = -2.89 kJ

Problem 2

For the decomposition of hydrogen peroxide, it is known that:

H2O2(l) → H2O(l) + 1/2 O2(g); ΔH = -98.2 kJ

Using this information, determine ΔH for the reaction:

2 H2O(l) + O2(g) → 2 H2O2(l)

Solution

When looking at the second equation, we see it is double the first reaction and in the opposite direction.

First change the direction of the first equation. When the direction of the reaction is changed, the sign

on ΔH changes for the reaction

H2O2(l) → H2O(l) + 1/2 O2(g); ΔH = -98.2 kJ

; becomes

H2O(l) + 1/2 O2(g) → H2O2(l); ΔH = +98.2 kJ

Second, multiply this reaction by 2. When multiplying a reaction by a constant, the ΔH is multiplied by

the same constant.

2 H2O(l) + O2(g) → 2 H2O2(l); ΔH = +196.4 kJ

Answer

ΔH = +196.4 kJ for the reaction: 2 H2O(l) + O2(g) → 2 H2O2(l)

Problem 3

Estimate the change in enthalpy, ΔH, for the following reaction:

H2 (g) + Cl2 (g) → 2 HCl (g)

Solution

To work this problem, think of the reaction in terms of simple steps:

Step 1 The reactant molecules, H2 and Cl2, break down into their atoms

H2(g) → 2 H(g)

Cl2(g) → 2 Cl(g)

Step 2 These atoms combine to form HCl molecules

2 H (g) + 2 Cl (g) → 2 HCl (g)

In the first step, the H-H and Cl-Cl bonds are broken. In both cases, one mole of bonds is broken. When

we look up the single bond energies for the H-H and Cl-Cl bonds, we find them to be +436 kJ/mol and +

243 kJ/mol, therefore for the first step of the reaction:

ΔH1 = +(436 kJ + 243 kJ) = +679 kJ

Bond breaking requires energy, so we expect the value for ΔH to be positive for this step.

In the second step of the reaction, two moles of H-Cl bonds are formed. Bond breaking liberates energy,

so we expect the ΔH for this portion of the reaction to have a negative value. Using the table, the single

bond energy for one mole of H-Cl bonds is found to be 431 kJ:

ΔH2 = -2(431 kJ) = -862 kJ

By applying Hess's Law, ΔH = ΔH1 + ΔH2

ΔH = +679 kJ - 862 kJ

ΔH = -183 kJ

Answer

The enthalpy change for the reaction will be ΔH = -183 kJ.

Problem 4

Find the amount of work done on the surroundings when 1 liter of an ideal gas, initially at a pressure of

10 atm, is allowed to expand at constant temperature to 10 liters by a) reducing the external pressure to

atm in a single step, b) reducing P first to 5 atm, and then to 1 atm, c) allowing the gas to expand into an

evacuated space so its total volume is 10 liters.

Solution: First, note that ΔV, which is a state function, is the same for each path:

V2 = (10/1) × (1 L) = 10 L, so ΔV = 9 L.

For path (a), w = –(1 atm)× (9 L) = –9 L-atm.

For path (b), the work is calculated for each stage separately:

w = –(1 atm) × (2–1 L) – (1 atm) × (10 L) = –9 L-atm

For path (c) the process would be carried out by removing all weights from the piston in Fig. 1 so that

the gas expands to 10 L against zero external pressure. In this case w = (0 atm) × 9 L = 0; that is, no work

is done because there is no force to oppose the expansion.

Problem 5

Hydrogen chloride gas readily dissolves in water, releasing 75.3 kJ/mol of heat in the process. If one

mole of HCl at 298 K and 1 atm pressure occupies 24.5 liters, find ΔU for the system when one mole of

HCl dissolves in water under these conditions.

Solution: In this process the volume of liquid remains practically unchanged, so ΔV = –24.5 L. The work

done is

w = –PΔV = –(1 atm)(–24.5 L) = 24.6 L-atm

(The work is positive because it is being done on the system as its volume decreases due to the

dissolution of the gas into the much smaller volume of the solution.) Using the conversion factor 1 L-atm

= 101.33 J mol–1

and substituting in Eq. 3 (above) we obtain

ΔU= q +PΔV = –(75300 J) + [101.33 J/L-atm) × (24.5 L-atm)] = –72.82 kJ

In other words, if the gaseous HCl could dissolve without volume change, the heat released by the

process (75.3 kJ) would cause the system’s internal energy to diminish by 75.3 kJ. But the disappearance

of the gaseous phase reduces the volume of the system. This is equivalent to compression of the system

by the pressure of the atmosphere performing work on it and consuming part of the energy that would

otherwise be liberated, reducing the net value of ΔU to –72.82 kJ.

Problem 6

The above reaction H2(g) + ½ O2(g) → H2O(g) is carried out at a constant pressure of 1 atm and a

constant temperature of 25° C. What quantity of heat q will cross the system boundary (and in which

direction?) For this reaction, the change in internal energy is ΔH = –240.59 kJ/mol.

Solution: The reaction itself (that is, the re-arrangement of the atoms from reactants to products

releases q = –240.59 kJ of heat. The work performed by the surroundings supplies an additional energy

of w = 1.24 kJ to the system. In order to maintain the constant 25° temperature, an equivalent quantity

of heat must be released: q = (–240.59 + 1.24) k J = –241.83 kJ

Problem 7

In determining the heat capacity of a calorimeter, a student mixes

100.0 g of water at 57.0 °C with 100.0 g of water, already in the calorimeter, at 24.2°C. (The specific heat

of water is 4.184 J g–1

K–1

)

After mixing and thermal equilibration with the calorimeter, the temperature of the water stabilizes at

38.7°C. Calculate the heat capacity of the calorimeter in J/K.

Solution:

The hot water loses heat, the cold water gains heat, and the calorimter itself gains heat, so this is

essentially a thermal balance problem. Conservation of energy requires that

qhot + qcold + qcal = 0

We can evaluate the first two terms from the observed temperature changes:

qhot = (100 g) (38.7 – 57.0) K (4.184 J g–1

K–1

) = –7657 J

qcold = (100 g) (38.7 – 24.2) K (4.184 J g–1

K–1

) = 6067 J

So qcal = (7657 – 6067) J = 1590 J

The calorimeter constantis (1590 J) / (38.7 – 24.2) K = 110 J K–1

Note: Strictly speaking, there is a fourth thermal balance term that must be considered in a highly

accurate calculation: the water in the calorimeter expands as it is heated, performing work on the

atmosphere.

Problem 8

A sample of biphenyl (C6H5)2 weighing 0.526 g was ignited in a bomb calorimeter initially at 25°C,

producing a temperature rise of 1.91 K. In a separate calibration experiment, a sample of benzoic acid

C6H5COOH weighing 0.825 g was ignited under identical conditions and produced a temperature rise of

1.94 K. For benzoic acid, the heat of combustion at constant pressure is known to be 3226 kJ mol–1

(that

is, ΔU° = –3226 kJ mol–1

.) Use this information to determine the standard enthalpy of combustion of

biphenyl.

Solution.The calorimeter constant is given by

The heat released by the combustion of the biphenyl at constant pressure is then Δ

(The negative sign indicates that heat is released in this process.) From the reaction equation

(C6H5

we have Δng = 12 - (19/2) = –5/2. Converting to Δ

ΔH = qV + ΔngRT

ΔH° = ΔU° – (5/2)(8.314 J mol–1

K–1

) =

This is the amount of heat that is lost by the system if reaction takes place at constant pressure and the

temperature is restored to its initial value.

The heat released by the combustion of the biphenyl at constant pressure is then ΔU:

(The negative sign indicates that heat is released in this process.) From the reaction equation

5)2(s) + 19/2 O2(g)→ 12 CO2(g) + 5 H2O(l)

5/2. Converting to ΔH, we substitute into

) = –6440 J mol–1

This is the amount of heat that is lost by the system if reaction takes place at constant pressure and the

re is restored to its initial value.

(The negative sign indicates that heat is released in this process.) From the reaction equation

This is the amount of heat that is lost by the system if reaction takes place at constant pressure and the