The rate constant for the forward and backward reactions of hydrolysis of ester are 1.1 × 10-2 and 1.5 × 10-3 min -1 respectively. The equilibrium constant of the reaction is

Question

The rate constant for the forward and backward reactions of hydrolysis of ester are 1.1 × 10-2 and 1.5 × 10-3 min -1 respectively. The equilibrium constant of the reaction is (A) 7.33 (B) 0.733 (C) 73.3 (D) 733

Tardigrade Admin · Accepted Answer

K =( K text forward / K text backward )=(1.1 × 10-2/1.5 × 10-3)=7.33

Q. The rate constant for the forward and backward reactions of hydrolysis of ester are $1.1 \times 1 0^{- 2}$ and $1.5 \times 1 0^{- 3} min^{- 1}$ respectively. The equilibrium constant of the reaction is

7.33

0.733

73.3

733

$K = \frac{K _{forward}}{K _{backward}} = \frac{1.1 \times 1 0 ^{- 2}}{1.5 \times 1 0 ^{- 3}} = 7.33$

Q. The rate constant for the forward and backward reactions of hydrolysis of ester are 1.1×10−2 and 1.5×10−3min−1 respectively. The equilibrium constant of the reaction is

7.33

0.733

73.3

733

K=Kbackward ​Kforward ​​=1.5×10−31.1×10−2​=7.33

Q. The rate constant for the forward and backward reactions of hydrolysis of ester are $1.1 \times 1 0^{- 2}$ and $1.5 \times 1 0^{- 3} min^{- 1}$ respectively. The equilibrium constant of the reaction is

$K = \frac{K _{forward}}{K _{backward}} = \frac{1.1 \times 1 0 ^{- 2}}{1.5 \times 1 0 ^{- 3}} = 7.33$