Question

${{K}_{p}}$ for a reaction at $25{}^\circ C$ is 10 atm. The activation energy for forward and reverse reactions are 12 and 20 kJ/mol respectively. The ${{K}_{c}}$for the reaction at $40{}^\circ C$will be
(A) $4.33\times {{10}^{-1}}M$
(B) $3.33\times {{10}^{-2}}M$
(C) $3.33\times {{10}^{-1}}M$
(D) $4.33\times {{10}^{-2}}M$

Answer 1

Hint: To attempt this question, first calculate the ${{K}_{c}}$ value at$25{}^\circ C$, and then calculate ${{K}_{c}}$for the reaction at $40{}^\circ C$. Use the Van't Hoff equation for the same.


Complete step by step solution:
Given,
Activation energy for forward reaction \[\left( {{E}_{f}} \right)=12kJ/mol\]
Activation energy for backward reaction \[\left( {{E}_{b}} \right)=20kJ/mol\]
${{K}_{p}}$at $25{}^\circ C$ = 10 atm and $\Delta n=1$.
Therefore, now calculate the value of ${{K}_{c}}$.
\[{{K}_{p}}={{K}_{c}}{{\left( RT \right)}^{\Delta n}}={{K}_{c}}\left( RT \right)\]
$\Rightarrow {{K}_{c}}=\dfrac{{{K}_{p}}}{RT}=\dfrac{10}{0.0821\times 298}=0.4M$
(Temperature values are placed after converting into kelvin, $25{}^\circ C+273=298K$ )
For calculating ${{K}_{c}}$ for the reaction use the vant’s Hoff equation:
     \[\log \dfrac{{{({{K}_{c}})}_{2}}}{{{({{K}_{c}})}_{1}}} =\dfrac{\Delta H}{2.303R}\left( \dfrac{{{T}_{2}}-{{T}_{1}}}{{{T}_{2}}{{T}_{1}}} \right)\]
Now, to calculate the value of $\Delta H$ we need the value of activation energy of forward and backward reaction.
$\Delta H={{E}_{f}}-{{E}_{b}}=12-20=-8kJ/mol=-8000J/mol$
\[\log \dfrac{{{({{K}_{c}})}_{313K}}}{{{({{K}_{c}})}_{298K}}} =\dfrac{-8000}{2.303\times 8.314}\left( \dfrac{313-298}{313\times 298} \right)\]
\[\log \dfrac{{{({{K}_{c}})}_{313K}}}{(0.4)} =\dfrac{-8000}{2.303\times 8.314}\left( \dfrac{15}{93870} \right)\]\[\]
\[\log \dfrac{{{({{K}_{c}})}_{313K}}}{(0.4)} =\dfrac{-120000}{1797342.22}\]
\[\log \dfrac{{{({{K}_{c}})}_{313K}}}{(0.4)} =-0.0667\]
Taking antilog both sides, we get
\[\dfrac{{{({{K}_{c}})}_{313K}}}{(0.4)} =1{{0}^{-0.0667}}\]
\[{{({{K}_{c}})}_{313K}} =0.4\times 0.8576=0.3430M=3.430\times 1{{0}^{-1}}M\]

So, the answer is option ©.

Note: While solving the question, remember to convert temperature from degree Celsius into Kelvin. Also, notice that the${{K}_{p}}$ value is given in atmosphere so to calculate ${{K}_{c}}$ use the value of universal gas constant (R) in atm not in Joules, i.e. 0.082 $L\cdot atm/K\cdot mol$not 8.314$J/K\cdot mol$.
Also, while calculating the${{K}_{c}}$ value at 313 K, either convert $\Delta H$ in Joules per mole or change gas constant (R) in $kJ/K\cdot mol$