The rate constant for forward reaction and backward reaction of hydrolysis of ester are 1.1 × 10-2 and 1.5 × 10-3 per minute respectively. Equilibrium constant for the reaction is CH 3 COOC 2 H 5+ H 2 O leftharpoons CH 3 COOH + C 2 H 5 OH

Question

The rate constant for forward reaction and backward reaction of hydrolysis of ester are 1.1 × 10-2 and 1.5 × 10-3 per minute respectively. Equilibrium constant for the reaction is CH 3 COOC 2 H 5+ H 2 O leftharpoons CH 3 COOH + C 2 H 5 OH (A) 33.7 (B) 7.33 (C) 5.33 (D) 33.3

Tardigrade Admin · Accepted Answer

kf=1.1 × 10-2, kb=1.5 × 10-3 Kc=(kf/kb)=(1.1 × 10-2/1.5 × 10-3)=7.33

Q. The rate constant for forward reaction and backward reaction of hydrolysis of ester are $1.1 \times 1 0^{- 2}$ and $1.5 \times 1 0^{- 3}$ per minute respectively.
Equilibrium constant for the reaction is
$C H_{3} COO C_{2} H_{5} + H_{2} O ⇌ C H_{3} COO H + C_{2} H_{5} O H$

33.7

7.33

5.33

33.3

$k_{f} = 1.1 \times 1 0^{- 2}, k_{b} = 1.5 \times 1 0^{- 3}$
$K_{c} = \frac{k _{f}}{k _{b}} = \frac{1.1 \times 1 0 ^{- 2}}{1.5 \times 1 0 ^{- 3}} = 7.33$

Q. The rate constant for forward reaction and backward reaction of hydrolysis of ester are 1.1×10−2 and 1.5×10−3 per minute respectively. Equilibrium constant for the reaction is CH3​COOC2​H5​+H2​O⇌CH3​COOH+C2​H5​OH