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Standard XII

Chemistry

Laws of Mass Conservation

One mole of N...

Question

One mole of $N_{2}$ and 3 moles of $H_{2}$ are mixed in a closed container of 2 litre capacity. At equilibrium if the container contains 1.2 moles of $N H_{3}$ , the active mass of $N_{2}$ is:

0.2

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0.6

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0.4

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1.2

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Solution

The correct option is A 0.2
0.6 moles of $N_{2}$ is used up to produce 1.2 mol of $N H_{3}$ , according to the equation,
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g), [N_{2}] = \frac{[1 - 0.6]}{2} = 0.2$

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One mole of N2 and 3 moles of H2 are mixed in a closed container of 2 litre capacity. At equilibrium if the container contains 1.2 moles of NH3, the active mass of N2 is:

The correct option is A 0.20.6 moles of N2 is used up to produce 1.2 mol of NH3, according to the equation, N2(g)+3H2(g)⇌2NH3(g),[N2]=[1−0.6]2=0.2

One mole of $N_{2}$ and 3 moles of $H_{2}$ are mixed in a closed container of 2 litre capacity. At equilibrium if the container contains 1.2 moles of $N H_{3}$ , the active mass of $N_{2}$ is:

The correct option is A 0.2
0.6 moles of $N_{2}$ is used up to produce 1.2 mol of $N H_{3}$ , according to the equation,
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g), [N_{2}] = \frac{[1 - 0.6]}{2} = 0.2$