In a reversible reaction, study of the mechanism says that both the forward and reverse reaction follows first order kinetics. If the half - life of forward reaction, ${⎛ ⎜ ⎝ t_{\frac{1}{2}} ⎞ ⎟ ⎠}_{f}$ is 400 sec and that of reverse reaction, ${⎛ ⎜ ⎝ t_{\frac{1}{2}} ⎞ ⎟ ⎠}_{f}$ is 250 sec. The equilbrium constant $(K_{e q})$ of the reaction is-
Given:
$⎛ ⎜ ⎝ t_{\frac{1}{2}} ⎞ ⎟ ⎠$ = $\frac{0.693}{k}$

1.6

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0.433

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0.625

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1.109

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Solution

The correct option is C 0.625
Given:
$⎛ ⎜ ⎝ t_{\frac{1}{2}} ⎞ ⎟ ⎠$ = $\frac{0.693}{k}$

Using this, we get:
$k_{f} = \frac{0.693}{400} s e c^{- 1}, K_{b} = \frac{0.693}{250} s e c^{- 1}$
So,
$K_{e q} = \frac{k_{f}}{k_{b}} = \frac{250}{400} = 0.625$ .

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Similar questions

Q. In a reversible reaction, study of the mechanism says that both the forward and reverse reaction follows first order kinetics. If the half - life of forward reaction,

{⎛ ⎜ ⎝ t_{\frac{1}{2}} ⎞ ⎟ ⎠}_{f}

is 400 sec and that of reverse reaction,

{⎛ ⎜ ⎝ t_{\frac{1}{2}} ⎞ ⎟ ⎠}_{f}

is 250 sec. The equilbrium constant

(K_{e q})

of the reaction is-
Given:

⎛ ⎜ ⎝ t_{\frac{1}{2}} ⎞ ⎟ ⎠

\frac{0.693}{k}

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The correct option is C 0.625Given: ⎛⎜⎝t12⎞⎟⎠=0.693k Using this, we get: kf=0.693400sec−1,Kb=0.693250sec−1 So, Keq=kfkb=250400=0.625.

The correct option is C 0.625
Given:
$⎛ ⎜ ⎝ t_{\frac{1}{2}} ⎞ ⎟ ⎠$ = $\frac{0.693}{k}$

Using this, we get:
$k_{f} = \frac{0.693}{400} s e c^{- 1}, K_{b} = \frac{0.693}{250} s e c^{- 1}$
So,
$K_{e q} = \frac{k_{f}}{k_{b}} = \frac{250}{400} = 0.625$ .