Question

The rate of disappearance of $A$ at two temperature for the equilibrium $A\rightleftharpoons B$ is given by
$-\frac{d\left[A\right]}{dt}=2\times {10}^{-2}\left[A\right]-4\times {10}^{-3}\left[B\right]$ at $300K$
$-\frac{d\left[A\right]}{dt}=4\times 106-2\left[A\right]-16\times {10}^{-4}\left[B\right]$ at $400K$
Calculate:
(i) equilibrium constants at $300K$ and $400K$ and
(ii) heat of reaction.

Answer 1

$A\rightleftharpoons B,K_{eq}=\frac{\left[B\right]}{\left[A\right]}$

At equilibrium, Rate of reaction becomes zero. Thus, at $300K$,

$\frac{-d\left[A\right]}{dt}=0\Rightarrow 2\times {10}^{-2}\left[A\right]-4\times {10}^{-3}\left[B\right]=0$

$\Rightarrow 4\times {10}^{-3}\left[B\right]=2\times {10}^{-2}\left[A\right]$

$\Rightarrow \frac{\left[B\right]}{\left[A\right]}=\frac{2\times {10}^{-2}}{4\times {10}^{-3}}$

at $300K\Rightarrow {K_{eq}}_1=\frac{\left[B\right]}{\left[A\right]}=5$

at $400K$

$2\times {10}^{-2}\left[A\right]-16\times {10}^{-4}\left[B\right]=0$

$\Rightarrow \frac{\left[B\right]}{\left[A\right]}=\frac{2\times {10}^{-2}}{16\times {10}^{-4}}$

${K_{eq}}_2=\frac{\left[B\right]}{\left[A\right]}=12.5$

Using Arrhenius Equation, we have

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

$\Rightarrow ln\left[\frac{12.5}{5}\right]=\frac{\Delta H}{R}[\frac{1}{300}-\frac{1}{400}]$

$\Rightarrow \Delta h=9.141KJ/mole$

This is the heat of Reaction.