Chapter 15, Problem 125AP

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

$\text{ln }K=\frac{-\Delta Hº}{RT}+C$

where C is a constant. The following table gives the equilibrium constant $\left(K_p\right)$ 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 $\Delta Hº$ for the reaction. (b) Use the van't Hoff equation to derive the following expression, which relates the equilibrium constants at two different temperatures:

$\text{ln}\frac{K_1}{K_2}=\frac{\Delta Hº}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

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 $\text{30°C}$ and 92.51 mmHg at $\text{50°C}$. Calculate the molar heat of vaporization of water.

Interpretation Introduction

Interpretation:

The $\Delta H^{\circ }$ 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)

$-115\text{kJ}/\text{mol}$

(b)

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

(c)

$43.4\text{kJ}/\text{mol}$

Explanation of Solution

a)The $\Delta H^{\circ }$ for the reaction graphically.

The $\Delta H^{\circ }$ for the reaction is calculated by plotting $\ln K_p$ versus $1/T$ asgiven below:

$\begin{array}{cc}\ln K_p& 1/T\\ 4.93& 0.00167\\ 1.63& 0.00143\\ -0.83& 0.00125\\ -2.77& 0.00111\\ -4.34& 0.00100\end{array}$

The slope of the plot is

$\text{Slope}=\Delta H^{\circ }/R$.

Here, $\Delta H^{\circ }$ is the standard enthalpy and $R$ is the gas constant.

Substitute the values of $\Delta H^{\circ }$ and $R$ in the above equation.

$\begin{array}{c}1.38\times {10}^4\cancel{\text{K}}=\frac{\Delta H^{\circ }}{8.314\text{J}/\text{mol}\text{.}\cancel{\text{K}}}\\ \Delta H^{\circ }=\left(1.38\times {10}^4\right)\left(8.314\text{J}/\text{mol}\right)\\ =-1.15\times {10}^5\text{J}/\text{mol}\\ =-115\text{kJ}/\text{mol}.\end{array}$

The $\Delta H^{\circ }$ for the reaction is $-115kJ/mol$.

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

$\ln \frac{K_1}{K_2}=\frac{-\Delta H^{\circ }}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$.

The vant Hoff’s equation at two different temperatures is

$\ln K_1=\frac{-\Delta H^o}{R{T}_1}+C$…… (1)

$\ln K_2=\frac{-\Delta H^o}{R{T}_2}+C$…… (2)

Here, $\Delta H^o$ is the standard enthalpy of the reaction, $K_1$, $K_2$ are the rate constants, $R$ is the gas constant, $T_1$ $T_2$ aretwo absolute temperatures, and $C$ is the molar concentration.

From (1) and (2):

$\begin{array}{c}\ln K_1-\ln K_2=\frac{-\Delta H^{\circ }}{R{T}_1}-\frac{-\Delta H^{\circ }}{R{T}_2}\\ \ln \frac{K_1}{K_2}=\frac{-\Delta H^{\circ }}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right).\end{array}$

An endothermic reaction has $\Delta H^{\circ }$ greater than zero, and temperature is also greater than zero. $\left(\Delta H^{\circ }>0\right)\left(T_2>T_1\right)$.

Then, temperature is

$\frac{\Delta H^{\circ }}{R}\left(\frac{\text{1}}{T_2}-\frac{1}{T_1}\right)<\text{0}$

$\ln \frac{K_1}{K_2}<0$

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

$\begin{array}{l}31.82mmHg\text{ at 30}°\text{C;}\\ \text{92}\text{.51mmHg at 50}°\text{C}\text{.}\end{array}$

From vant’s Hoff equation,

$\ln \frac{K_1}{K_2}=\frac{\Delta H^{\circ }}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$.

Here, $\frac{K_1}{K_2}$ is the rate constant for the reaction, $\Delta H^o$ is the standard enthalpy, $R$ is the gas constant, and $T_2,T_1$ are the absolute temperatures.

Substitute the values of $\frac{K_1}{K_2}$, $\Delta H^{\circ }$, $\text{R,}$ and $T_2,T_1$ in the above equation.

$\begin{array}{c}\ln \frac{31.82\text{ mmHg}}{92.51\text{ mmHg}}=\frac{\Delta H^{\circ }}{8.314\text{ J}/\text{mol}\text{.K}}\left(\frac{1}{323}-\frac{1}{303}\right)\\ \ln 0.3439=\frac{\Delta H^{\circ }}{8.314\text{ J}/\text{mol}\text{.K}}\left(\frac{1}{323}-\frac{1}{303}\right)\\ =\frac{\Delta H^{\circ }}{8.314\text{J}/\text{mol}\text{.K}}\left(\frac{-20}{97869}\right)\\ -1.0674=\frac{\Delta H^{\circ }}{8.314\text{J}/\text{mol}\text{.K}}\left(-2.0435\times {10}^{-4}\right).\end{array}$

On further solving

$\begin{array}{c}-8.8743=\Delta H^{\circ }\left(-2.0435\times {10}^{-4}\right)\\ \Delta H^{\circ }=4.3426\times {10}^4\text{J}/\text{mol}\\ =43.4\text{}\text{}\text{kJ}/\text{mol}.\end{array}$

The molar heat ofvaporization of water is $43.4\text{kJ}/\text{mol}$.