Consider the following equilibrium reactions at 25°C, A (g )+ 3 B (g ) 2C (g ),Keq = x A(g) + D ( g ) 2E ( g) ,Keq = y B(g) + D (g ) F(g),Keq = z then the value of equilibrium constant for the reaction 2C(g) + 4D(g) 2E(g) + 3F(g), at 25°C is

Question

Consider the following equilibrium reactions at 25°C, A (g )+ 3 B (g ) <=> 2C (g ),Keq = x A(g) + D ( g ) <=> 2E ( g) ,Keq = y B(g) + D (g ) <=> F(g),Keq = z then the value of equilibrium constant for the reaction 2C(g) + 4D(g) <=> 2E(g) + 3F(g), at 25°C is (A) (yz3/x) (B) (xy/z3) (C) (xz/y) (D) (y2z/x)

Tardigrade Admin · Accepted Answer

2C (g) leftharpoons A(g)+3B(g) K1=(1/x) ldots(i) A(g)+D(g) leftharpoons 2E(g) K2=Y ldots(ii) 3B (g)+3D(g) leftharpoons 3F(g) K3=z3 ldots(iii) Adding (i), (ii) and (iii) 2C(g)+4D(g) leftharpoons 2E(g)+3F(g), K=(Yz3/X)

Q. Consider the following equilibrium reactions at $2 5^{\circ} C$ ,
$A (g) + 3 B (g) 2 C (g), K_{e q} = x$
$A (g) + D (g) 2 E (g), K_{e q} = y$
$B (g) + D (g) F (g), K_{e q} = z$
then the value of equilibrium constant for the reaction
$2 C (g) + 4 D (g) 2 E (g) + 3 F (g)$ , at $2 5^{\circ} C$ is

$\frac{y z ^{3}}{x}$

$\frac{x y}{z ^{3}}$

$\frac{x z}{y}$

$\frac{y ^{2} z}{x}$

$2 C (g) ⇌ A (g) + 3 B (g) K_{1} = \frac{1}{x} \dots (i)$
$A (g) + D (g) ⇌ 2 E (g) K_{2} = Y \dots (ii)$
$3 B (g) + 3 D (g) ⇌ 3 F (g) K_{3} = z^{3} \dots (iii)$
Adding $(i)$ , $(ii)$ and $(iii)$
$2 C (g) + 4 D (g) ⇌ 2 E (g) + 3 F (g), K = \frac{Y z ^{3}}{X}$

Q. Consider the following equilibrium reactions at 25∘C, A(g)+3B(g)​2C(g),Keq​=x A(g)+D(g)​2E(g),Keq​=y B(g)+D(g)​F(g),Keq​=z then the value of equilibrium constant for the reaction 2C(g)+4D(g)​2E(g)+3F(g), at 25∘C is

xyz3​

z3xy​

yxz​

xy2z​