Question

$K_{p}$ for the reaction $\underset{(s)}{2 BaO_{2}} \rightleftharpoons \underset{(s)}{2BaO}+O_{2}(g)$ is $1.6 \times 10^{-4}$ atm at $400^{\circ} C$. Heat of reaction is $-25.14 \,kcal$ 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)

$\log \frac{ K _{ P _{2}}}{ K _{ p _{1}}}=\frac{\Delta H }{2.303 R }\left(\frac{1}{ T _{1}}-\frac{1}{ T _{2}}\right)$

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

$K _{ p _{2}}=1.46 \times 10^{-5} \,atm$

$K _{ p _{2}}= P _{ O _{2}}=1.46 \times 10^{-5} \,atm$

Apply, $PV = n\,RT$

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

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

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