Question

The given reaction is $2A\to P$, which of the equilibrium concentration correspond to $K_c$ value for 0.02 moles of $P$ and 0.1 moles of $A$ at equilibrium?

• A. 0.04, 0.2
• B. 0.18, 0.3
• C. 0.08, 0.2
• D. 0.16, 0.8

Answer 1

Answer: (B), (C)

The correct options are
B 0.18, 0.3
C 0.08, 0.2

For the reaction $2A\to P$; $K_c=\frac{\left[P\right]}{[A{]}^2}$

where [P] = 0.02 & [A] = 0.1

Therefore, $K_c$ = $\frac{0.02}{0.1\times 0.1}$ = 2

A) $K_c$ = $\frac{0.04}{0.2\times 0.2}$ = 1

B) $K_c$ = $\frac{0.18}{0.3\times 0.3}$ = 2

C) $K_c$ = $\frac{0.08}{0.2\times 0.2}$ = 2

D) $K_c$ = $\frac{0.16}{0.8\times 0.8}$ = 0.25

Therefore B & C are correct.