Question

$k_p$ for a reaction at ${25}^{0}C$ in 10 atm. Then $K_c$ for the reaction at ${40}^{0}C$ will be:

(Given :$\Delta H=-8kJ$ and antilog (0.067) = 0.1167)

• A. $4.33\times {10}^{-1}M$
• B. $3.33\times {10}^{-2}M$
• C. $3.33\times {10}^{-1}M$
• D. None of these

Answer 1

The correct option is C $3.33\times {10}^{-1}M$

$K_P$ at ${25}^{0}C$ is $10$ atm

$K_P$ at ${40}^{0}C$

$In\left(\frac{K_{{40}^{0}C}}{K_{{25}^{0}C}}\right)=-\left(\frac{\Delta H}{R}\right)(\frac{1}{T^{{40}^{0}C}}-\frac{1}{T^{{25}^{0}C}})$

$In\left(\frac{K}{10}\right)=-0.1547=log\left(\frac{K}{10}\right)=-0.067$

$\frac{10}{K}=1.167$

$K=8.569$

SInce $K$ units are ${atm}^1$, So $\Delta n=1$

$K_P=K_C{\left(RT\right)}^1$

$K_C=\frac{8.569}{0.082\times 313}=0.33$