Chemistry - Chemical Equilibrium Question with Solution | TestHub

Let's analyze the equilibrium:

$$N_2(g)+2H_2(g)\rightleftharpoons N_2{H}_4(g)$$

Initial mass ratio of $N_2$ and $H_2$ is 7:1.

Molar mass of $N_2$ = 28 g/mol

Molar mass of $H_2$ = 2 g/mol

Let the initial mass of $N_2$ be 7x and the initial mass of $H_2$ be x.

Initial moles of $N_2$ = $\frac{7x}{28}=\frac{x}{4}$

Initial moles of $H_2$ = $\frac{x}{2}$

Let P be the total pressure at equilibrium, and $P_{N_2{H}_4}=0.2P$.

| | $N_2$ | $2H_2$ | $N_2{H}_4$ |

|-------------|------------|------------|------------|

| Initial | $x/4$ | $x/2$ | 0 |

| Change | -y | -2y | +y |

| Equilibrium | $x/4-y$ | $x/2-2y$ | y |

Total moles at equilibrium = $(x/4-y)+(x/2-2y)+y=x/4+x/2-2y=3x/4-2y$

Partial pressure of $N_2{H}_4$ = (moles of $N_2{H}_4$ / total moles) * Total pressure

$0.2P=\frac{y}{\frac{3x}{4}-2y}\ast P$

$0.2=\frac{y}{\frac{3x}{4}-2y}$

$0.2(\frac{3x}{4}-2y)=y$

$\frac{0.6x}{4}-0.4y=y$

$0.15x=1.4y$

$y=\frac{0.15x}{1.4}=\frac{15x}{140}=\frac{3x}{28}$

Moles of $N_2$ at equilibrium = $\frac{x}{4}-\frac{3x}{28}=\frac{7x-3x}{28}=\frac{4x}{28}=\frac{x}{7}$

Moles of $H_2$ at equilibrium = $\frac{x}{2}-2(\frac{3x}{28})=\frac{x}{2}-\frac{6x}{28}=\frac{14x-6x}{28}=\frac{8x}{28}=\frac{2x}{7}$

Partial pressure of $N_2$ = (moles of $N_2$ / total moles) * Total pressure

$P_{N_2}=\frac{\frac{x}{7}}{\frac{3x}{4}-2(\frac{3x}{28})}\ast P=\frac{\frac{x}{7}}{\frac{3x}{4}-\frac{3x}{14}}\ast P=\frac{\frac{x}{7}}{\frac{21x-6x}{28}}\ast P=\frac{\frac{x}{7}}{\frac{15x}{28}}\ast P=\frac{x}{7}\ast \frac{28}{15x}\ast P=\frac{4}{15}P$

Partial pressure of $H_2$ = (moles of $H_2$ / total moles) * Total pressure

$P_{H_2}=\frac{\frac{2x}{7}}{\frac{15x}{28}}\ast P=\frac{2x}{7}\ast \frac{28}{15x}\ast P=\frac{8}{15}P$

Comparing $P_{N_2}$ and $P_{H_2}$:

$2P_{N_2}=2\ast \frac{4}{15}P=\frac{8}{15}P=P_{H_2}$

Therefore, $2P_{N_2}=P_{H_2}$

Option C is correct.

Final Answer: The final answer is $\boxed{C}$