Consider the following reaction equilibrium

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

Initially, 1 mole of $N_{2}$ and 3 moles of $H_{2}$ are taken in a 2 L flask. At equilibrium state if, the number of moles of $N_{2}$ is 0.6, what is the total number of moles of all gases present in the flask?

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TestHub · Accepted Answer

Number of moles of N 2 = 0.6 = 1 - x ∴ x = 1 - 0.6 = 0.4 So, 3 - 3 x = 3 - 3 × 0.4 = 1.8 , and 2 x = 2 × 0.4 = 0.8 Therefore, the total number of moles at equilibrium. = 1 - x + 3 - 3 x + 2 x = 0.6 + 1.8 + 0.8 = 3.2 mol

Chemistry - Chemical Equilibrium Question with Solution | TestHub

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