A mixture of `NH_(3(g))` and `N_(2)H_(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))rarrN_(2(g))+3H_(2(g))`
`N_(2)H_(4_((g)))rarrN_(2(g))+2H_(2(g))`
After decomposition is complete, the total pressure at `1200 K` is found to be `4.5 atm`. Find the amount (mole) per cent of `N_(2)H_(4(g))` in the original mixture.

Options

A. 0.2
B. 0.25
C. 0.5
D. 0.75

Explanation

To solve the problem, we need to analyze the decomposition of the gases involved and how they contribute to the total pressure after heating. Let's break it down step by step.

### Step 1: Define Variables
Let:
- \( N_1 \) = number of moles of \( NH_3 \)
- \( N_2 \) = number of moles of \( N_2H_4 \)

### Step 2: Initial Conditions
At 300 K, the total pressure \( P \) is given as 0.5 atm. According to the ideal gas law, we can express the total number of moles in terms of pressure:
\[ P = \frac{nRT}{V} \]
Where \( n \) is the total number of moles, \( R \) is the gas constant, and \( V \) is the volume. Thus:
\[ 0.5 = \frac{(N_1 + N_2)RT}{V} \]

### Step 3: Decomposition Reactions
The decomposition reactions are:
1. \( 2NH_3 \rightarrow N_2 + 3H_2 \)
2. \( N_2H_4 \rightarrow N_2 + 2H_2 \)

From these reactions, we can derive the number of moles produced after decomposition:
- From \( N_1 \) moles of \( NH_3 \), the moles of \( N_2 \) produced will be \( \frac{N_1}{2} \) and the moles of \( H_2 \) produced will be \( \frac{3N_1}{2} \).
- From \( N_2 \) moles of \( N_2H_4 \), the moles of \( N_2 \) produced will be \( N_2 \) and the moles of \( H_2 \) produced will be \( 2N_2 \).

### Step 4: Total Moles After Decomposition
After decomposition, the total number of moles at 1200 K will be:
\[ \text{Total moles} = \left(\frac{N_1}{2} + N_2\right) + \left(\frac{3N_1}{2} + 2N_2\right) \]
This simplifies to:
\[ \text{Total moles} = 2N_1 + 3N_2 \]

### Step 5: Pressure at 1200 K
At 1200 K, the total pressure is given as 4.5 atm. Using the ideal gas law again:
\[ 4.5 = \frac{(2N_1 + 3N_2)RT}{V} \]

### Step 6: Set Up the Equations
Now we have two equations:
1. From the initial condition:
\[ 0.5 = \frac{(N_1 + N_2)RT}{V} \quad \text{(1)} \]
2. From the final condition:
\[ 4.5 = \frac{(2N_1 + 3N_2)RT}{V} \quad \text{(2)} \]

### Step 7: Divide the Equations
Dividing equation (2) by equation (1):
\[ \frac{4.5}{0.5} = \frac{(2N_1 + 3N_2)}{(N_1 + N_2)} \]
This simplifies to:
\[ 9 = \frac{(2N_1 + 3N_2)}{(N_1 + N_2)} \]

### Step 8: Cross Multiply
Cross multiplying gives:
\[ 9(N_1 + N_2) = 2N_1 + 3N_2 \]
Expanding and rearranging:
\[ 9N_1 + 9N_2 = 2N_1 + 3N_2 \]
\[ 7N_1 + 6N_2 = 0 \quad \text{(3)} \]

### Step 9: Solve for Ratios
From equation (3):
\[ 7N_1 = -6N_2 \]
This implies:
\[ \frac{N_2}{N_1} = \frac{7}{6} \]

### Step 10: Calculate Mole Percent
To find the mole percent of \( N_2H_4 \):
Let \( N_1 = 6x \) and \( N_2 = 7x \). The total moles \( N_{total} = N_1 + N_2 = 6x + 7x = 13x \).
The mole percent of \( N_2H_4 \) is:
\[ \text{Mole percent of } N_2H_4 = \frac{N_2}{N_{total}} \times 100 = \frac{7x}{13x} \times 100 = \frac{7}{13} \times 100 \approx 53.85\% \]

### Final Answer
The mole percent of \( N_2H_4 \) in the original mixture is approximately **53.85%**.