For the reaction 2 NH 3 ( g ) ⟶ N 2 ( g ) + 3 H 2 ( g ) , what is the % of NH 3 converted if the mixture diffuses twice as fast as that of SO 2 under similar conditions.

Question

For the reaction 2 NH 3 ​ ( g ) ⟶ N 2 ​ ( g ) + 3 H 2 ​ ( g ) , what is the % of NH 3 converted if the mixture diffuses twice as fast as that of SO 2 under similar conditions.

TestHub · Accepted Answer

Here's a brief solution:

Using Graham's Law of Diffusion:

$\frac{r _{mi x}}{r _{S O_{2}}} = \frac{M _{S O_{2}}}{M _{mi x}}$

Given that the mixture diffuses twice as fast as SO₂, we have:

$\frac{2}{1} = \frac{64}{M _{mi x}}$

$4 = \frac{64}{M _{mi x}}$

$M_{mi x} = 16$

Now, consider the reaction:

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

Let $x$ be the moles of N₂ formed. Then $3 x$ is the moles of H₂ formed, and $2 x$ is the moles of NH₃ consumed. If we start with 1 mole of NH₃, then $(1 - 2 x)$ moles of NH₃ remain.

Total moles at equilibrium = $(1 - 2 x) + x + 3 x = 1 + 2 x$

$M_{mi x} = \frac{T o t a l ma ss}{T o t a l m o l es}$

$16 = \frac{1 \times 17}{1 + 2 x}$

$16 + 32 x = 17$

$32 x = 1$

$x = \frac{1}{32}$

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