Question

# For the reaction $$2NH_3(g) \rightarrow N_2(g) + 3H_2(g)$$, what is the $$\%$$ of $$NH_3$$ converted if the mixture diffuses twice as fast as that of $$SO_2$$ under similar conditions :

A
3.125%
B
31.25%
C
6.25%
D
62.5%

Solution

## The correct option is D $$6.25\%$$  $$\displaystyle NH_3$$$$\displaystyle N_2$$ $$\displaystyle H_2$$Initial moles  10 0  Equilibrium moles $$\displaystyle 1-2x$$$$\displaystyle x$$ $$\displaystyle 3x$$  Equilibrium mole fraction $$\displaystyle \frac {1-2x}{1+2x}$$ $$\displaystyle \frac {x}{1+2x}$$ $$\displaystyle \frac {3x}{1+2x}$$The rate of diffusion is inversely proportional to the square root of molar mass.The mixture diffuses twice as fast as that of $$SO_2$$ under similar conditions.$$\displaystyle \frac {r}{r'} = \sqrt {\frac {M'}{M}}$$ $$\displaystyle \frac {2}{1} = \sqrt {\frac {64}{M}}$$ $$\displaystyle M = 16$$  $$g/mol$$$$\displaystyle 16 = \frac {17(1-2x) + 28(x) + 2(3x)}{1+2x}$$$$\displaystyle 16+32x = 17-34x+28x+6x$$$$\displaystyle 32x = 1$$$$\displaystyle x = 0.03125$$The percentage of ammonia converted$$\displaystyle \frac {2x}{1} \times 100 = \frac {2 \times 0.03125}{1} \times 100 = 6.25$$%Chemistry

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