If the pressure of `N_(2)//H_(2)` mixture in a closed vessel is 100 atmospheres and `20%` of the mixture reacts then the pressure at the same temperature would be .

Options

A. Same as original
B. 110 atm
C. 90 atm
D. 80 atm

Answer

To solve the problem, we need to determine the pressure of the nitrogen and hydrogen mixture after a certain percentage of it has reacted. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Reaction
The reaction between nitrogen (N₂) and hydrogen (H₂) is given as:
\[ \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \]

### Step 2: Initial Conditions
- The initial pressure of the mixture (P₁) is 100 atmospheres.
- The initial number of moles of the gases can be considered as:
- N₂: 1 mole
- H₂: 3 moles
- Total initial moles (n₁) = 1 + 3 = 4 moles.

### Step 3: Determine the Degree of Reaction
Given that 20% of the mixture reacts, we can denote the degree of reaction as α = 0.2.

### Step 4: Calculate Moles at Equilibrium
At equilibrium:
- Moles of N₂ = \( 1 - \alpha = 1 - 0.2 = 0.8 \) moles
- Moles of H₂ = \( 3 - 3\alpha = 3 - 3(0.2) = 2.4 \) moles
- Moles of NH₃ produced = \( 2\alpha = 2(0.2) = 0.4 \) moles

### Step 5: Total Moles at Equilibrium
Total moles at equilibrium (n₂):
\[ n₂ = \text{Moles of N₂} + \text{Moles of H₂} + \text{Moles of NH₃} \]
\[ n₂ = 0.8 + 2.4 + 0.4 = 3.6 \text{ moles} \]

### Step 6: Use the Ideal Gas Law
Since the volume (V) and temperature (T) remain constant, we can use the ideal gas law:
\[ P_1V = n_1RT \]
\[ P_2V = n_2RT \]

### Step 7: Relate Initial and Final Pressures
Dividing the two equations:
\[ \frac{P_1}{P_2} = \frac{n_1}{n_2} \]

### Step 8: Substitute Known Values
Substituting the known values:
- \( P_1 = 100 \) atm
- \( n_1 = 4 \) moles
- \( n_2 = 3.6 \) moles

This gives us:
\[ \frac{100}{P_2} = \frac{4}{3.6} \]

### Step 9: Solve for P₂
Rearranging the equation to solve for \( P_2 \):
\[ P_2 = 100 \times \frac{3.6}{4} \]
\[ P_2 = 90 \text{ atm} \]

### Conclusion
The pressure at the same temperature after 20% of the mixture reacts is 90 atmospheres.