You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XIII

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

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

0.433

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

0.625

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

1.109

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

Open in App

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

Suggest Corrections

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}

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

(t_{1 / 2})

f is

400

s and that of reverse reaction

(t_{1 / 2})

b is

250

s, the equilibrium constant of the reaction is:

Q. For an equilibrium reaction involving gases, the forward reaction is first order while the reverse reaction is second order. The units of

K_{p}

for forward equilibrium is:

In an exothermic reaction $A \to B$ , the activation energy of reverse reaction is twice that of forward reaction. If enthalpy of the reaction is -80 kJ $m o l^{- 1}$ , the activation energy of the reverse reaction is :

Q. For an equilibrium change involving gaseous phase, the forward reaction is first order while the reverse reaction is second order. The unit of

K_{p}

for the forward equilibrium is :

Join BYJU'S Learning Program

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