Question

For the first order opposed by first order reaction ;

k_f = 9 × 10^–3 min^–1 and k_b = 2 × 10^–3 min^–1

If we start with the concentration of A equal to 1 (M) what will be concentration of B in 10³/11 mins?

• A. 0.818 (M)
• B. 0.409 (M)
• C. 0.517(M)
• D. 0.190 (M)

Answer 1

Answer: (C)

0.517(M)

Answer 2

The integrated rate equation is

ln $\frac{x_{eq}}{(x_{eq} - x)}$ = (k_f + k_b) × t

ln $\frac{x_{eq}}{(x_{eq} - x)}$ = 11 × 10^–3 × $\frac{10^{3}}{11}$= 1 = ln e

\ $\frac{x_{eq}}{(x_{eq} - x)}$ = e or x_eq = e x_eq – x e

or x e = e x_eq – x_eq

or x = $\frac{x_{eq}(e - 1)}{e}$

or x = x_eq $\frac{(2.718 - 1)}{2.718}$ = x_eq × 0.632

$\mathrm { A } \underset { \mathrm { k } _ { \mathrm { b } } } { \stackrel { \mathrm { k } _ { \mathrm { f } } } { \longleftrightarrow } } \mathrm { B }$ 1 0

(1 – x_eq) x_eq

\ $\frac{x_{eq}}{1 - x_{eq}}$ = $\frac{k_{f}}{k_{b}}$ = $\frac{9}{2}$

or 2 x_eq = 9 – 9 x_eq

or x_eq = $\frac{9}{11}$ = 0.818

\ x = 0.818 × 0.632

or x = 0.517 (M)