The rate constant for forward 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 - CH3COOC2H5 - H2O- CH3COOH-C2H5OH

The correct option is D 7.33Rate constant of forward reaction (K_f)=1.1× 10^ 2 min^ 1 Rate constant of backward reaction (K_b)=1.5× 10^ 3 min^ 1 # ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Order of Reaction

The rate cons...

Question

The rate constant for forward and backward reaction of hydrolysis of ester are $1.1 \times 10^{- 2}$ and $1.5 \times 10^{- 3}$ per minute respectively.

Equilibrium constant for the reaction is :

$C H_{3} C O O C_{2} H_{5} + H_{2} O ⇌ C H_{3} C O O H + C_{2} H_{5} O H$

4.33

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5.33

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6.33

No worries! We‘ve got your back. Try BYJU‘S free classes today!

7.33

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 7.33
Rate constant of forward reaction $(K_{f}) = 1.1 \times 10^{- 2} m i n^{- 1}$
Rate constant of backward reaction $(K_{b}) = 1.5 \times 10^{- 3} m i n^{- 1}$
Equilibrium constant $(K_{c})$ $= \frac{K_{f}}{K_{b}}$

$= \frac{1.1 \times 10^{- 2}}{1.5 \times 10^{- 3}}$

$= 7.33$
Hence, option $(D)$ is correct.

Suggest Corrections

Similar questions

Q. The rate constant for forward reaction and backward reaction of hydrolysis of ester are

1.1 \times 10^{- 2}

and

1.5 \times 10^{- 3}

per minute respectively. Equilibrium constant for the reaction is :

C H_{3} C O O C_{2} H_{5} + H_{2} O ⇌ C H_{3} C O O H + C_{2} H_{5} O H

Q. The rate constant for the forward and backward reactions of hydrolysis of ester are

1.1 \times 10^{- 2}

and

1.5 \times 10^{- 3}

respectively. The equilibrium constant of the following reaction is:

C H_{3} C O O C_{2} H_{5} + H^{+} ⇌ C H_{3} C O O H + C_{2} H_{5} O H

Q. The rate constant for forward and backward reaction of hydrolysis of ester are

1.1 \times 10^{- 2}

and

1.5 \times 10^{- 3}

per minute respectively. Equilibrium constant for the reaction is?

The rate constant for the forward and backward reactions of the hydrolysis of ester are

1.1 \times 10^{- 2}

and

1.5 \times 10^{- 3}

min^-1respectively. The equilibrium constant of the reaction is:

Q. The rate constants for forward and backward reactions of hydrolysis of ethyl acetate are

1.1 \times 10^{- 2} a n d 1.5 \times 10^{- 3} m i n^{- 1}

respectively. The equilibrium constant for the reaction is

Join BYJU'S Learning Program

The rate constant for forward 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 :CH3COOC2H5+H2O ⇌ CH3COOH+C2H5OH

The correct option is D 7.33Rate constant of forward reaction (Kf)=1.1×10−2 min−1Rate constant of backward reaction (Kb)=1.5×10−3 min−1 Equilibrium constant (Kc)=KfKb=1.1×10−21.5×10−3=7.33Hence, option (D) is correct.

The rate constant for forward and backward reaction of hydrolysis of ester are $1.1 \times 10^{- 2}$ and $1.5 \times 10^{- 3}$ per minute respectively.

Equilibrium constant for the reaction is :

$C H_{3} C O O C_{2} H_{5} + H_{2} O ⇌ C H_{3} C O O H + C_{2} H_{5} O H$