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Q.

For the reaction
$2 {NH}_{3} (g) ⟶ N_{2} (g) + 3 H_{2} (g)$ ,
what is the % of NH₃ converted if the mixture diffuses twice as fast as that of SO₂ under similar conditions.

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a

3.125 %

b

6.25 %

c

31.25 %

d

62.5 %

answer is C.

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Detailed Solution

$\frac{r_{mix}}{r_{{SO}_{2}}} = \sqrt{\frac{M_{{SO}_{2}}}{M_{mix}}} = \sqrt{\frac{64}{M_{mix}}} = 2$

M_mix=16


Initial number of moles	2	0	0
Equilibrium number of moles	2−2x	x	3x
Mole fraction	$\frac{2 - 2 x}{2 + 2 x}$	$\frac{x}{2 + 2 x}$	$\frac{3 x}{2 + 2 x}$

$\frac{17 (2 - 2 x) + 28 x + 6 x}{2 + 2 x} = 16$

x= $\frac{1}{16}$
The percentage of ammonia converted is $\frac{2 \times 100}{2 \times 16}$ =6.25%

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