Chemistry
Chemistry
4th Edition
ISBN: 9780078021527
Author: Julia Burdge
Publisher: McGraw-Hill Education
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Chapter 15, Problem 125AP

The dependence of the equilibrium constant of a reaction on temperature is given by the van't Hoff equation:

ln  K = Δ H º R T + C

where C is a constant. The following table gives the equilibrium constant ( K p ) for the reaction at various temperatures:

KP

138

5.12

0.436

0.0626

0.0130

TOO

600

700

800

900

1000

(a) Determine graphically the Δ H º for the reaction. (b) Use the van't Hoff equation to derive the following expression, which relates the equilibrium constants at two different temperatures:

ln K 1 K 2 = Δ H º R ( 1 T 2 1 T 1 )

How does this equation support the prediction based on Le Châ�telier's principle about the shift in equilibrium with temperature? (c) The vapor pressures of water are 31.82 mmHg at 30°C and 92.51 mmHg at 50°C . Calculate the molar heat of vaporization of water.

Expert Solution & Answer
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Interpretation Introduction

Interpretation:

The ΔH for the reaction is to be calculated, and the derivation of the given expression at two different temperatures, the prediction of the equation based on Le Chatelier’s, and the molar heat of vaporization of waterareto be determined.

Concept Introduction:

Achemical equilibrium is a state of the chemical reaction when the rate of forward reaction becomes equal to the rate of reverse reaction and the concentrations of the products and reactants become constant, which are known as equilibrium concentrations.

When energy is released in a reaction, the reaction is called exothermic, and the reactions in which energy is absorbed are endothermic reactions.

According to Le Chatelier’s Principle, when a system at equilibrium is subjected to any change, the system tries to undo the effect of that change by shifting its equilibrium in the desired direction.

The van't Hoff equation provides information about the temperature dependence of the equilibrium constant.

Answer to Problem 125AP

Solution:

(a)

115 kJ/mol

(b)

According to Le Chatelier’s principles, increase in temperature favors the forward endothermic reaction, and decrease in temperature favors an exothermic reaction.

(c)

43.4kJ/mol

Explanation of Solution

a)The ΔH for the reaction graphically.

The ΔH for the reaction is calculated by plotting lnKp versus 1/T asgiven below:

lnKp1/T4.930.001671.630.001430.830.001252.770.001114.340.00100

Chemistry, Chapter 15, Problem 125AP

The slope of the plot is

Slope=ΔH/R.

Here, ΔH is the standard enthalpy and R is the gas constant.

Substitute the values of ΔH and R in the above equation.

1.38×104K=ΔH8.314 J/mol.KΔH=(1.38×104)(8.314 J/mol)=1.15×105 J/mol=115 kJ/mol.

The ΔH for the reaction is 115 kJ/mol.

b)The following equation support the prediction based on Le-Chatelier’s principle about the shift in equilibrium with temperature

lnK1K2=ΔHR(1T21T1).

The vant Hoff’s equation at two different temperatures is

lnK1=ΔHoRT1+C…… (1)

lnK2=ΔHoRT2+C…… (2)

Here, ΔHo is the standard enthalpy of the reaction, K1, K2 are the rate constants, R is the gas constant, T1 T2 aretwo absolute temperatures, and C is the molar concentration.

From (1) and (2):

lnK1lnK2=ΔHRT1ΔHRT2lnK1K2=ΔHR(1T21T1).

An endothermic reaction has ΔH greater than zero, and temperature is also greater than zero. (ΔH>0)(T2>T1).

Then, temperature is

ΔHR(1T21T1)<0

lnK1K2<0

In a reversible reaction at equilibrium, when the equilibrium constant is more, the products are also more or the rate constant of forward reaction is more, and hence the products are more. According to LeChatelier’s principle, an increase in temperature favors endothermic reaction, and a decrease in temperature favors exothermic reaction.

c) Molar heat of vaporization of water

The vapor pressure of water is

31.82mmHg at 30°C;92.51mmHg at 50°C.

From vant’s Hoff equation,

lnK1K2=ΔHR(1T21T1).

Here, K1K2 is the rate constant for the reaction, ΔHo is the standard enthalpy, R is the gas constant, and T2,T1 are the absolute temperatures.

Substitute the values of K1K2, ΔH, R, and T2,T1 in the above equation.

ln31.82 mmHg92.51 mmHg=ΔH8.314 J/mol.K(13231303)ln0.3439=ΔH8.314 J/mol.K(13231303)=ΔH8.314 J/mol.K(2097869)1.0674=ΔH8.314 J/mol.K(2.0435×104).

On further solving

8.8743=ΔH(2.0435×104)ΔH=4.3426×104 J/mol=43.4 kJ/mol.

The molar heat ofvaporization of water is 43.4 kJ/mol.

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Chapter 15 Solutions

Chemistry

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