Chapter 11, Problem 11.2P

Interpretation Introduction

Interpretation:

The heat of reaction has to be calculated from the given equilibrium conversion for the elementary reaction.

Explanation of Solution

For elementary reaction $\text{A}\stackrel{}{\rightleftharpoons }\text{B}$,

Equilibrium conversion,

  ${\text{X}}_{\text{e}}\text{ = }\frac{{\text{K}}_{\text{c}}}{1+{\text{K}}_{\text{c}}}$

Therefore,

  ${\text{K}}_{\text{c}}\text{ = }\frac{{\text{X}}_{\text{e}}}{1-{\text{X}}_{\text{e}}}$

Where ${\text{K}}_{\text{c}}$ is the concentration equilibrium constant.

Condition 1:

At temperature $127°\text{C}$, equilibrium conversion is 0.8

  $\begin{array}{l}{\text{T}}_{\text{1}}=127°\text{C (or) 400 K}\\ {\text{K}}_{\text{c}}{}_{\text{1}}=\frac{0.8}{1-0.8}\\ =\text{4}\end{array}$

Condition 2:

At temperature $227°\text{C}$, equilibrium conversion is 0.5

  $\begin{array}{l}{\text{T}}_2=227°\text{C (or) 500 K}\\ {\text{K}}_{\text{c}}{}_{\text{1}}=\frac{0.5}{1-0.5}\\ =1\end{array}$

  $\begin{array}{l}\text{ln}\frac{{\text{K}}_{\text{c1}}}{{\text{K}}_{\text{c2}}}=\frac{\Delta {\text{H}}_{\text{RX}}}{\text{R}}\left[\frac{1}{{\text{T}}_{\text{1}}}-\frac{1}{{\text{T}}_2}\right]\\ \therefore \Delta {\text{H}}_{\text{RX}}=\frac{{\text{T}}_{\text{2}}{\text{T}}_{\text{1}}}{{\text{T}}_{\text{2}}-{\text{T}}_{\text{1}}}\text{R ln}\frac{{\text{K}}_{\text{c1}}}{{\text{K}}_{\text{c2}}}\mathrm{......}(1)\end{array}$

Where,

$\Delta {\text{H}}_{\text{RX}}$ is hear of reaction.

${\text{T}}_{\text{1}}{\text{ and T}}_{\text{2}}$  is temperature.

R is gas constant ($1.987{\text{ cal K}}^{-1}{\text{ mol}}^{-1})$

${\text{K}}_{\text{c1}}\text{ and}{\text{K}}_{\text{c2}}$ are concentration equilibrium constants.

Substitute ${\text{T}}_{\text{1}}{\text{ = 400 K, T}}_2{\text{ = 500 K; K}}_{\text{c1}}{\text{ =4, K}}_{\text{c2}}\text{ = 1}$ in equation (1).

  $\begin{array}{l}\Delta {\text{H}}_{\text{RX}}=\frac{500\times 400}{500-400}\text{K}\times \text{1}\text{.987}\frac{\text{cal}}{\text{K mol}}\times \text{ln}\frac{4}{1}\\ =\frac{200000}{100}\text{K}\times \text{1}\text{.987}\frac{\text{cal}}{\text{K mol}}\times 1.386\\ =2000\times 1.987\times 1.386\\ =5509\frac{\text{cal}}{\text{mol}}\end{array}$

Thus, the heat of reaction when equilibrium conversion is 0.8 at $127°\text{C}$ and 0.5 at $227°\text{C}$ is $5509\text{ cal/mol}\text{.}$