Equilibrium Question 749
Question: 8 mol of $ AB_3(g) $ are introduced into a $ 1.0dm^{3} $ vessel. If it dissociates as $ 2AB_3(g)\rightarrow A_2(g)+3B_2(g) $ . At equilibrium, 2 mol of $ A_2 $ are found to be present. The equilibrium. constant of this reaction is
Options:
A) 2
B) 3
C) 27
D) 36
Show Answer
Answer:
Correct Answer: C
Solution:
$ \begin{aligned} & \begin{matrix} {} & {} & {} & 2AB_3(g)\rightarrow A_2(g)+3B_2(g) \\ \end{matrix} \\ & \begin{matrix} at,t=0 & {} & 8 & \begin{matrix} {} & {} & 0 & \begin{matrix} {} & \begin{matrix} {} & 0 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ & \begin{matrix} at,eq. & (8-2\times 2) & {} & \begin{matrix} 2 & \begin{matrix} {} & 3\times 2 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ & \begin{matrix} {} & {} & {} & \begin{matrix} =4 & {} & {} & \begin{matrix} \begin{matrix} 2 & {} & {} \\ \end{matrix} & 6 \\ \end{matrix} \\ \end{matrix} \\ \end{matrix} \\ \end{aligned} $ now $ K_{C}=\frac{[ A_2 ]{{[ B_2 ]}^{3}}}{{{[ AB_3 ]}^{2}}}=\frac{2/1\times {{[6/1]}^{3}}}{{{[4/1]}^{2}}}=27 $
BETA
