Question

For the reaction, ${{2N}}{{{H}}_{{3}}}\left( {{g}} \right) \to {{{N}}_{{2}}}\left( {{g}} \right){{ + 3}}{{{H}}_{{2}}}\left( {{g}} \right)$, what is the percentage of ${{N}}{{{H}}_{{3}}}$ converted, if the mixture diffuses twice fast as that of ${{S}}{{{O}}_2}$ under similar conditions?
A) $3.125\,\%$
B) $31.25\,\%$
C) $6.25\,\%$
D) None of these

Answer 1

Diffusion is the process of the movement of the substance from the high concentration to the lower concentration. In the case of the diffusion of the gases, gas molecules are moved from the higher concentration to the lower gas concentration. The mass of the gas molecules, temperature affect the diffusion of the gases. The higher the mass lower is the rate of diffusion and vice versa. An increase in the temperature increases the kinetic energy of the gas molecules leads to an increase in the rate of diffusion.

Formula Used: The mathematical expression of Graham’s Law is given as follows:
$\dfrac{{{{{R}}_{{1}}}\,}}{{{{{R}}_{{2}}}}}{{ = }}\dfrac{{\sqrt {{{{M}}_{{2}}}} }}{{\sqrt {{{{M}}_{{1}}}} }}$

Complete step-by-step answer:
The mathematical representation of Graham’s Law of diffusion is given as follows:
${{R}}\,{{\alpha }}\dfrac{1}{{\sqrt {{M}} }}$
If there are two gases one with the rate of diffusion is ${{{R}}_1}$, and its molecular weight is ${{{M}}_1}$, the rate of diffusion of another gas is ${{{R}}_2}$, and its molecular weight is ${{{M}}_2}$.
$\dfrac{{{{{R}}_{{1}}}\,}}{{{{{R}}_{{2}}}}}{{ = }}\dfrac{{\sqrt {{{{M}}_{{2}}}} }}{{\sqrt {{{{M}}_{{1}}}} }}$
Here, in the question, the rate of the diffusion of the gas mixture is twice that of the rate of diffusion of ${{S}}{{{O}}_2}$. The molecular weight of the gas ${{S}}{{{O}}_2}$ is ${{64}}\,{{g}}\,{{mo}}{{{l}}^{{{ - 1}}}}$.

Now, use Graham’s law of diffusion,
$\Rightarrow\dfrac{{{{{R}}_{{{mix}}}}\,}}{{{{{R}}_{{{S}}{{{O}}_2}}}}}{{ = }}\dfrac{{\sqrt {{{{M}}_{{{S}}{{{O}}_2}}}} }}{{\sqrt {{{{M}}_{{{mix}}}}} }}$
$\Rightarrow {{{R}}_{{{mix}}}}{{ = 2}}{{{R}}_{{{S}}{{{O}}_2}}}$ and ${{{M}}_{{{S}}{{{O}}_2}}}{{ = 64}}\,{{g}}\,{{mo}}{{{l}}^{{{ - 1}}}}$
$\Rightarrow \dfrac{{{{2}} \times {{{R}}_{{{S}}{{{O}}_2}}}\,}}{{{{{R}}_{{{S}}{{{O}}_2}}}}}{{ = }}\dfrac{{\sqrt {64} }}{{\sqrt {{{{M}}_{{{mix}}}}} }}$
$\Rightarrow \sqrt {{{{M}}_{{{mix}}}}} {{ = }}\dfrac{{\sqrt {64} }}{2}$
$\Rightarrow \sqrt {{{{M}}_{{{mix}}}}} {{ = }}4$
$\Rightarrow {{{M}}_{{{mix}}}}{{ = }}16$

Here, we can see that the molecular weight of the mixture obtained is 16 units.

Now, here to determine the percentage of the ammonia converted used the balanced chemical reaction given in the question.
${{2N}}{{{H}}_{{3}}}\left( {{g}} \right) \to {{{N}}_{{2}}}\left( {{g}} \right){{ + 3}}{{{H}}_{{2}}}\left( {{g}} \right)$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{2N}}{{{H}}_{{3}}}\left( {{g}} \right) \to {{{N}}_{{2}}}\left( {{g}} \right)\,\,\,\,{{ + }}\,\,\,{{3}}{{{H}}_{{2}}}\left( {{g}} \right)$

Initial moles	2	0	0	total
Equilibrium moles	2-2x	x	3x
Total moles	2-2x+	x+	3x+	=2+2x
Mole fraction	$\dfrac{{{{2 - 2x}}}}{{{{2 + 2x}}}}$	$\dfrac{{{x}}}{{{{2 + 2x}}}}$	$\dfrac{{{{3x}}}}{{\,{{2 + 2x}}\,}}$

The molar mass of the mixture is equal to the sum of the product of the molar mass of each component and the mole fraction of that component.

$\Rightarrow \dfrac{{17\left( {{{2 - 2x}}} \right){{ + 28}}\left( {{x}} \right){{ + 2}}\left( {{{3x}}} \right)}}{{{{2 + 2x}}}} = 16$
$\Rightarrow \left( {{{34 - 34x}}} \right){{ + }}\left( {{{28x}}} \right){{ + }}\left( {{{6x}}} \right) = 16\left( {{{2 + 2x}}} \right)$
$\Rightarrow {{x}} = \dfrac{1}{{16}}$

Here, moles of ammonia converted is given as 2x therefore the percentage of the ammonia converted is determined by taking the ratio of moles converted into the products to the initial moles.

$\Rightarrow {{percentage}}\,{{of}}\,{{ammonia}}\,\,{{conveted}}\,{{ = }}\dfrac{{\,{{moles}}\,{{converted}}\,\,{{into}}\,{{products}}}}{{{{Initial}}\,\,{{moles}}}} \times {{100}}\,{{\% }}$
$\Rightarrow {{percentage}}\,{{of}}\,{{ammonia}}\,\,{{conveted}}\,{{ = }}\dfrac{{{{2x}}}}{{{2}}} \times {{100}}\,{{\% }}$
Now, here substitute the value of x.
$\Rightarrow {{percentage}}\,{{of}}\,{{ammonia}}\,\,{{conveted}}\,{{ = }}\dfrac{{{2}}}{{{2}}} \times \dfrac{1}{{16}} \times {{100}}\,{{\% }}$
$\Rightarrow {{percentage}}\,{{of}}\,{{ammonia}}\,\,{{conveted}}\,{{ = }}\,6.25\,{{\% }}$

Thus, the percentage of ammonia converted is $6.25\,{{\% }}$. Therefore, option (C) is the correct answer to the given question.

Note: In the case of the diffusion of the gases, Graham’s Law of diffusion is used. This law was given by Thomas Graham when he has studied the behavior of gases. As per Graham’s law, the rate of gas diffusion is inversely proportional to the square root of the density or molecular weight of the gas.