The rate constant for forward and backward reactions of hydrolysis of ester are 1.1×10−2 and 1.5×10−3 per minute respectively. Equilibrium constant for the reaction is

Equilibrium constant K=kbkf =Backward rate constantForward rate constant K=1.5×10−31.1×10−2=1.51.1×10=1511=7.33

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The rate constant for forward and backward reactions of hydrol

Question

The rate constant for forward and backward reactions 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

4.33

5.33

6.33

7.33

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Emily Discussed

The rate constant for forward and backward reactions 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

14 mins ago

Discuss this question LIVE

14 mins ago

Equilibrium constant

K = \frac{k _{f}}{k _{b}}

= \frac{Forward rate constant}{Backward rate constant}

K = \frac{1.1 \times 10 ^{- 2}}{1.5 \times 10 ^{- 3}} = \frac{1.1 \times 10}{1.5} = \frac{11}{15} = 7.33

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Question Text	The rate constant for forward and backward reactions 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
Answer Type	Text solution:1
Upvotes	150