Question

Variation of equilibrium constant $K$ with temperature $T$ is given by van't Hoff equation:
$\log K=\log A-\frac{\Delta H^o}{2.303RT}$
A graph between $\log K$ and $T^{-1}$ was a straight line as shown in the figure and having $\theta ={\tan }^{-1}(-0.5)$ and $OP=10$. Calculate:
(a) $\Delta H^o$ (Standard heat of reaction) when $T=298K$
(b) $A$ (pre-exponential factor)
(c) Equilibrium constant $K$ at $298K$
(d) $K$ at $798K$, if $\Delta H^o$ is independent of temperature

Answer 1

$Explanation:$

$a)lo{g}_{10}K=lo{g}_{10}A-\frac{\Delta H^o}{2.303RT}$

It is an equation of a straight line of the $y=c+mx$

Slope $mtan\theta =$ $\frac{\Delta H^o}{2.303R}$

$0.5=\frac{\Delta H^0}{2.303}\times 8.314$

$\Delta H^o=9.574Jmo{l}^{-1}$

$b)Intercept,D^{{}^{\prime }}=lo{g}_{10}A=10A={10}^{10}$

$D^{{}^{\prime }}={10}^{10}$

$c)logK=10-\frac{9.574}{2.303\times 8.314\times 298}$

$K=9.96\times {10}^9$

$d)log\frac{K_2}{K_1}=\frac{\Delta H}{20303}\times R(\frac{1}{T_1}-\frac{1}{T_2})$

$\frac{log\left(K_2\right)}{9.96\times {10}^9}=1.05\times {10}^{-3}$

$log\left(K_2\right)=9.98\times {10}^6$

$K_2=e^{9.98\times {10}^6}$