Consider the following equations for a cell reaction. A + B leftharpoons C + D , E °=x volt, K eq = K 1 2 A +2 B leftharpoons 2 C +2 D , E °=y volt, K eq = K 2 Then,

Question

Consider the following equations for a cell reaction. A + B leftharpoons C + D , E °=x volt, K eq = K 1 2 A +2 B leftharpoons 2 C +2 D , E °=y volt, K eq = K 2 Then, (A) x=y, K 1= K 2 (B) x=y, K12=K2 (C) x=2 y, K 1=2 K 2 (D) x2=y, K 12= K 2

Tardigrade Admin · Accepted Answer

When a chemical reaction is multiplied/divided EMF of the changed equation remains constant but equilirbium constant is raised to power of that change.

Thus,

x = y

A + B \to C + D,

K_{1} = \frac{[ C ] [ D ]}{[ A ] [ B ]}

2 A + 2 B \to 2 C + 2 D,

K_{2} = \frac{[ C ] ^{2} [ D ] ^{2}}{[ A ] ^{2} [ B ] ^{2}} = K_{1}^{2}

Q. Consider the following equations for a cell reaction.
$A + B ⇌ C + D, E^{\circ} = x$ volt, $K_{e q} = K_{1}$
$2 A + 2 B ⇌ 2 C + 2 D, E^{\circ} = y$ volt, $K_{e q} = K_{2}$
Then,

$x = y, K_{1} = K_{2}$

$x = y, K_{1}^{2} = K_{2}$

$x = 2 y, K_{1} = 2 K_{2}$

$x^{2} = y, K_{1}^{2} = K_{2}$

Q. Consider the following equations for a cell reaction. A+B⇌C+D,E∘=x volt, Keq​=K1​ 2A+2B⇌2C+2D,E∘=y volt, Keq​=K2​ Then,

x=y,K1​=K2​

x=y,K12​=K2​

x=2y,K1​=2K2​

x2=y,K12​=K2​

When a chemical reaction is multiplied/divided EMF of the changed equation remains constant but equilirbium constant is raised to power of that change. Thus, x=y A+B→C+D, K1​=[A][B][C][D]​ 2A+2B→2C+2D, K2​=[A]2[B]2[C]2[D]2​=K12​

Q. Consider the following equations for a cell reaction.
$A + B ⇌ C + D, E^{\circ} = x$ volt, $K_{e q} = K_{1}$
$2 A + 2 B ⇌ 2 C + 2 D, E^{\circ} = y$ volt, $K_{e q} = K_{2}$
Then,

$x = y, K_{1} = K_{2}$

$x = y, K_{1}^{2} = K_{2}$

$x = 2 y, K_{1} = 2 K_{2}$

$x^{2} = y, K_{1}^{2} = K_{2}$