Question

For the reaction N2 + O2 \xrightleftharpoons{{}} 2NO KC is 200 then, KP for reaction 2NO \xrightleftharpoons{{}} N2 + O2 at constant temperature :

• A. 5 × 10–3
• B. 2.5 × 10–3
• C. 2 × 102
• D. 2 × 10–2

Answer 1

Answer: (A)

5 × 10–3

Answer 2

The reaction $N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$ has $\Delta n = (2) - (1+1) = 0$. For reactions where $\Delta n = 0$, $K_p = K_c$. Thus, for the forward reaction, $K_p = K_c = 200$. The reverse reaction is $2NO(g) \rightleftharpoons N_2(g) + O_2(g)$. The equilibrium constant for the reverse reaction ($K_c'$) is the reciprocal of the equilibrium constant for the forward reaction ($K_c$), so $K_c' = \frac{1}{K_c} = \frac{1}{200}$. For the reverse reaction, $\Delta n' = (1+1) - (2) = 0$. Since $\Delta n' = 0$, $K_p'$ for the reverse reaction is equal to $K_c'$ for the reverse reaction. Therefore, $K_p' = K_c' = \frac{1}{200}$. Converting this to scientific notation: $\frac{1}{200} = \frac{1}{2 \times 100} = 0.5 \times 10^{-2} = 5 \times 10^{-3}$.