Question

A mixture of $NH_3(g)$ and $N_2H_4(g)$ is placed in a sealed container at 300 K. The total pressure is 0.5 atm. The container is heated to 1200 K at which time both substances decompose completely according to the equations

$2NH_3(g) \longrightarrow N_2(g) + 3H_2(g)$

and

$N_2H_4(g) \longrightarrow N_2(g) + 2H_2(g)$

After decomposition is complete, the total pressure at 1200 K is found to be 4.5 atm. Find the mole% of $N_2H_4$ in the original mixture.

• A. 20%
• B. 25%
• C. 50%
• D. 75%

Answer 1

Answer: (B)

25%

Answer 2

Let initial moles of NH₃ and N₂H₄ be x and y.

• Initial total moles: n = x + y.
• Initial pressure at 300 K: P₁ ∝ n·300.
• On decomposition at 1200 K: • NH₃ → produces 2 moles gas per mole NH₃ ⇒ 2x
  • N₂H₄ → produces 3 moles gas per mole N₂H₄ ⇒ 3y
  Total final moles = 2x + 3y.
• Final pressure at 1200 K: P₂ ∝ (2x+3y)·1200.
  Since volume and R cancel,

$$\frac{P_2}{P_1}=\frac{(2x+3y)\,1200}{(x+y)\,300} =\frac{4\,(2x+3y)}{(x+y)}=\frac{4.5}{0.5}\times\frac{0.5}{0.5}=9?$$

More directly:

$$P_1=0.5\text{ atm},\quad P_2=4.5\text{ atm}\;\Rightarrow\;\frac{P_2}{P_1}=9$$ $$\frac{(2x+3y)\,1200}{(x+y)\,300}=9 \;\Longrightarrow\;\frac{2x+3y}{x+y}= \frac{9}{4}=2.25$$ $$2x+3y=2.25(x+y)\;\Longrightarrow\;2x+3y=2.25x+2.25y \;\Longrightarrow\;0.75y=0.25x\;\Longrightarrow\;y=\frac{x}{3}$$

Mole fraction of N₂H₄ = y/(x+y) = (x/3)/(4x/3) = 1/4 = 25%.