For $N_{2} O_{4}$ (g) $⇋$ $2 N O_{2}$ (g),the value of degree of dissociation as a function of equilibrium pressure P and equilibrium constant $K_{p}$ is $α = \sqrt{\frac{K_{p}}{(4 p + K_{p})}}$ True or false?

True

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False

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Solution

The correct option is A
True

Assume there are initially n moles of reactant only. Derive $K_{p}$ as in the previous problem

$K_{p} = \frac{P^{2} N O_{2}}{P N_{2} O_{4}} = \frac{[{2 α l (1 + α)} P]^{2}}{{(1 - α) / (1 + α)} P} = \frac{4 α^{2} P}{1 - α^{2}}$

Solving for $α$ , we get $α = \sqrt{\frac{K_{p}}{(4 P + K_{p})}}$

The same could be tried for other reactions as well:

$2 N H_{3} (g) ⇋ N_{2} (g) + 3 H_{2} (g)$

$2 N O (g) ⇋ N_{2} (g) + O_{2} (g)$

$P C l_{5} (g) ⇋ P C l_{3} (g) + C l_{2} (g)$

$2 H l (g) ⇋ H_{2} (g) + l_{2} (g)$

$2 H_{2} S (g) ⇋ 2 H_{2} (g) + S_{2} (g)$

What about the following reaction?

$N H_{4} H S (s) ⇋ N H_{3} (g) + H_{2} S (g)$

Hint within a hint: Active mass of a solid

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Similar questions

Q. For the dissociation reaction

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g),

the degree of dissociation

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Q. For the reaction

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For N2O4(g)⇋2NO2(g),the value of degree of dissociation as a function of equilibrium pressure P and equilibrium constant Kp is α=√Kp(4p+Kp) True or false?

For $N_{2} O_{4}$ (g) $⇋$ $2 N O_{2}$ (g),the value of degree of dissociation as a function of equilibrium pressure P and equilibrium constant $K_{p}$ is $α = \sqrt{\frac{K_{p}}{(4 p + K_{p})}}$ True or false?