Question

A mixture of NH3(g) and N2H4 (g) is placed in a sealed container at 300 K. The pressure within the container is 0.6 atm. When the container is heated to 1000 K, where the two gases undergo decomposition reactions as given below

2NH3(g) → N2(g) + 3H2(g) and N₂H₄(g) → N2(g) + 2H2(g)

The pressure of the container after complete decomposition becomes 4.8 atm. The mole percent of NH3(g) in the original mixture was x. The value of $\frac{x}{10}$ is

Answer 1

6

Answer 2

Let $n_{NH_3}$ and $n_{N_2H_4}$ be the initial number of moles of NH3(g) and N2H4(g) in the sealed container. The initial total number of moles is $n_{initial} = n_{NH_3} + n_{N_2H_4}$. The initial pressure is $P_{initial} = 0.6$ atm at $T_{initial} = 300$ K. The final temperature is $T_{final} = 1000$ K, and the final pressure is $P_{final} = 4.8$ atm. The volume of the container $V$ is constant.

Using the ideal gas law, $PV = nRT$. For the initial state: $P_{initial} V = n_{initial} R T_{initial}$ $0.6 V = (n_{NH_3} + n_{N_2H_4}) R (300)$ (1)

When the container is heated to 1000 K, the gases decompose completely according to the reactions:

1. $2NH_3(g) \rightarrow N_2(g) + 3H_2(g)$ From this reaction, 2 moles of NH3 produce $1+3=4$ moles of gas. Thus, 1 mole of NH3 produces 2 moles of gas. If initially there were $n_{NH_3}$ moles of NH3, they will produce $2n_{NH_3}$ moles of gas after decomposition.

2. $N_2H_4(g) \rightarrow N_2(g) + 2H_2(g)$ From this reaction, 1 mole of N2H4 produces $1+2=3$ moles of gas. If initially there were $n_{N_2H_4}$ moles of N2H4, they will produce $3n_{N_2H_4}$ moles of gas after decomposition.

The total number of moles after complete decomposition is $n_{final} = 2n_{NH_3} + 3n_{N_2H_4}$. For the final state: $P_{final} V = n_{final} R T_{final}$ $4.8 V = (2n_{NH_3} + 3n_{N_2H_4}) R (1000)$ (2)

Divide equation (2) by equation (1): $\frac{4.8 V}{0.6 V} = \frac{(2n_{NH_3} + 3n_{N_2H_4}) R (1000)}{(n_{NH_3} + n_{N_2H_4}) R (300)}$ $8 = \frac{(2n_{NH_3} + 3n_{N_2H_4}) \times 1000}{(n_{NH_3} + n_{N_2H_4}) \times 300}$ $8 = \frac{10 (2n_{NH_3} + 3n_{N_2H_4})}{3 (n_{NH_3} + n_{N_2H_4})}$ $24 (n_{NH_3} + n_{N_2H_4}) = 10 (2n_{NH_3} + 3n_{N_2H_4})$ $24n_{NH_3} + 24n_{N_2H_4} = 20n_{NH_3} + 30n_{N_2H_4}$ $4n_{NH_3} = 6n_{N_2H_4}$ $2n_{NH_3} = 3n_{N_2H_4}$

We are given that the mole percent of NH3(g) in the original mixture was x. Mole percent of NH3 = $\frac{n_{NH_3}}{n_{NH_3} + n_{N_2H_4}} \times 100 = x$.

From the relationship $2n_{NH_3} = 3n_{N_2H_4}$, we can express $n_{N_2H_4}$ in terms of $n_{NH_3}$: $n_{N_2H_4} = \frac{2}{3} n_{NH_3}$.

Substitute this into the mole percent expression: $x = \frac{n_{NH_3}}{n_{NH_3} + \frac{2}{3} n_{NH_3}} \times 100$ $x = \frac{n_{NH_3}}{(1 + \frac{2}{3}) n_{NH_3}} \times 100$ $x = \frac{n_{NH_3}}{(\frac{3+2}{3}) n_{NH_3}} \times 100$ $x = \frac{n_{NH_3}}{(\frac{5}{3}) n_{NH_3}} \times 100$ $x = \frac{3}{5} \times 100$ $x = 60$

The question asks for the value of $\frac{x}{10}$. $\frac{x}{10} = \frac{60}{10} = 6$.