The value of K c us 64 at 800 K for the reaction N 2 g + 3 H 2 g → 2 NH 3 g The value of K c for the following reaction is : NH 3 g → 1 2 N 2 g + 3 2 H 2 ( g )

N 2 g + 3 H 2 g ⇌ 2 NH 3 g K C = NH 3 2 N 2 H 2 3 = 64 For the reaction NH 3 g ⇌ 1 2 N 2 g + 3 2 H 2 g K ' C = N 2 1 / 2 H 2 3 / 2 NH 3 = 1 K C = 1 64 = 1 8

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ChemistryChemical EquilibriumProblems involving KcEasy2 minPYQ_2020

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The value of $K_{c} us 64 at 800 K$ for the reaction

$N_{2} (g) + 3 H_{2} (g) \to 2 {NH}_{3} (g)$

The value of $K_{c}$ for the following reaction is :

${NH}_{3} (g) \to \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g)$

Options:

Answer:

Solution:

$N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$

$K_{C} = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}} = 64$

For the reaction

${NH}_{3} (g) ⇌ \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g)$

$K'_{C} = \frac{{[N_{2}]}^{1 / 2} {[H_{2}]}^{3 / 2}}{[{NH}_{3}]} = \frac{1}{\sqrt{K_{C}}} = \frac{1}{\sqrt{64}} = \frac{1}{8}$

Stream:JEESubject:ChemistryTopic:Chemical EquilibriumSubtopic:Problems involving Kc

⏱ 2mℹ️ Source: PYQ_2020

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