NH3 (calculated vapour density 8.5) partly decomposes by NH3(g) 1/2 N2(g) + 3/2 H2(g). The vapour density of the equilibrium mixture is 6.8. The degree of dissociation of NH3 is

JEE Advanced levelSingle correct optionDifficulty 3 of 3ce203

$NH X_{3}$ (calculated vapour density $8.5$ ) partly decomposes by $NH X_{3} (g) \frac{1}{2} N X_{2} (g) + \frac{3}{2} H X_{2} (g)$ . The vapour density of the equilibrium mixture is $6.8$ . The degree of dissociation of $NH X_{3}$ is

One mole of

NH X_{3}

gives

\frac{1}{2} + \frac{3}{2} = 2

mol of products, so

n = 2

and total moles are

1 + α

. Hence

α = (D - d) / d = (8.5 - 6.8) /6.8 = 0.250