Question

$K_p$ for the reaction $\underset{(s)}{2Ba{O}_2}\rightleftharpoons \underset{(s)}{2BaO}+O_2(g)$ is $1.6\times 10^{-4}$ atm at $400^{\circ }C$. Heat of reaction is $-25.14kcal$ What will be the number of moles of $O_2$ gas produced at $500^{\circ }C$ temperature. The reaction is carried in 2 litre vessel

• A. $0.022\times 10^{-10}$
• B. $10^{-8}$
• C. $10^{-24}$
• D. $0.046\times 10^{-5}$

Answer 1

Answer: (D)

$\log \frac{K_{P_2}}{K_{p_1}}=\frac{\Delta H}{2.303R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

$\log \frac{K_{p_2}}{1.6\times 10^{-4}}=\frac{-25.14}{2.303\times 2\times 10^{-3}}\frac{(773-673)}{773\times 673}$

$K_{p_2}=1.46\times 10^{-5}atm$

$K_{p_2}=P_{O_2}=1.46\times 10^{-5}atm$

Apply, $PV=nRT$

$1.46\times 10^{-5}\times 2=n\times 0.0821\times 773$

$n=\frac{1.46\times 10^{-5}\times 2}{0.0821\times 2773}=\frac{2.92\times 10^{-5}}{63.46}$

$=0.046\times 10^{-5}$