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 # ...

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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

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5.33

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6.33

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7.33

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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.

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Similar questions

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

1.1 \times 10^{- 2}

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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 :

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