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 2 - f is 400 sec and that of reverse reaction- t 1-2 f is 250 secThe equilbrium constant K eq of the reaction is-Given-B- 0-433C- 0-625D- 1-109

The correct option is C 0.625Given: ( t_12 )=0.693k Using this, we get: k_f = 0.693400 sec^ 1, K_b = 0.693250 sec^ 1 So, K_eq = k_fk_b = 250400 = 0.625. ...

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Standard XII

Chemistry

Derivation of Kp and Kc

In a reversib...

Question

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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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,
$K_{e q} = \frac{k_{f}}{k_{b}} = \frac{250}{400} = 0.625$ .