Question

Consider the following equilibrium reactions at $25^{\circ}C$,
$A (g )+ 3 B (g ) \ce{<=>} 2C (g ),K_{eq} = x$
$A(g) + D ( g ) \ce{<=>} 2E ( g) ,K_{eq} = y$
$B(g) + D (g ) \ce{<=>} F(g),K_{eq} = z$
then the value of equilibrium constant for the reaction
$ 2C(g) + 4D(g) \ce{<=>} 2E(g) + 3F(g)$, at $25^{\circ}C$ is

• A. $\frac{yz^3}{x}$
• B. $\frac{xy}{z^3}$
• C. $\frac{xz}{y}$
• D. $\frac{y^2z}{x}$

Answer 1

Answer: (A)

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