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Compressibility Factor

NH3 converts ...

Question

${N H}_{3}$ converts into $N_{2}$ and $H_{2}$ according to the following equation:
${2 N H}_{3} \to N_{2} + {3 H}_{2}$
$x % o f {N H}_{3}$ is converted into product. Rate of diffusion of gaseous mixture is double than that of ${S O}_{2}$ under identical conditions. Calculate $x$ .

A

98.23 %

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B

31.25 %

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C

25.27 %

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D

6.25 %

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Solution

The correct option is C $6.25 %$
According to the Graham's law of diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its Molar mass.

$%$ of $N H_{3}$ converted into product = $x$ .
$2 N H_{3} \to N H_{2} + 3 H_{2}$
$t = 0$ $2$ $0$ $0$
$t = e q u i l i b r i u m$ $2 - 2 x$ $x$ $3 x$

Total number of moles = $2 - 2 x + x + 3 x$

$M_{m i x} = \frac{(n_{N H_{3}}) \times M_{N H_{3}} + (n_{H_{2}}) \times M_{H_{2}} + (n_{H_{2}}) \times M_{H_{2}}}{T o t a l n u m b e r o f m o l e s}$

$M_{m i x} = \frac{34}{2 + 2 x}$

Now, $\frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$

$r_{S O_{2}} = \frac{1}{2} r_{m i x}$

$\frac{r_{S O_{2}}}{r_{m i x}} = \sqrt{\frac{M_{m i x}}{M_{S O_{2}}}}$
$M_{M i x} = 16$

Hence, $16 = \frac{17}{1 + x}$

$x = \frac{1}{16}$
$%$ of $x = 6.25 %$
Hence, the correct option is D.

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