A reaction `X_2(g)rarr Z(g) +1/2Y(g)` exhibits an increase in pressure from 150 mm to 170 mm in 10 minutes. The rate of disappearance of `X_2` in mm per minute is

Options

A. 2 mm per minute
B. 8 mm per minute
C. 4 mm per minute
D. 6 mm per minute

Answer

To solve the problem, we need to determine the rate of disappearance of \(X_2\) in mm per minute based on the change in pressure over a given time period.

### Step-by-Step Solution:

1. **Identify Initial and Final Pressures:**
- Initial pressure (\(P_i\)) = 150 mm
- Final pressure (\(P_f\)) = 170 mm

2. **Calculate the Change in Pressure:**
\[ \Delta P = P_f - P_i = 170 \, \text{mm} - 150 \, \text{mm} = 20 \, \text{mm} \]

3. **Understand the Reaction:**
The reaction is given as:
\[ X_2(g) \rightarrow Z(g) + \frac{1}{2}Y(g) \]
For every 1 mole of \(X_2\) that disappears, 1 mole of \(Z\) and 0.5 moles of \(Y\) are produced. This means that the total increase in pressure due to the products can be calculated.

4. **Relate Pressure Change to Moles:**
The increase in pressure is due to the formation of \(Z\) and \(Y\). The total change in pressure from the reaction can be expressed as:
- Pressure change due to \(Z\) = \(P\)
- Pressure change due to \(\frac{1}{2}Y\) = \(\frac{P}{2}\)
- Therefore, total pressure change = \(P + \frac{P}{2} = \frac{3P}{2}\)

5. **Set Up the Equation:**
We know that the total increase in pressure (\(\Delta P\)) is 20 mm:
\[ \frac{3P}{2} = 20 \, \text{mm} \]

6. **Solve for \(P\):**
\[ 3P = 40 \, \text{mm} \implies P = \frac{40}{3} \approx 13.33 \, \text{mm} \]

7. **Calculate the Rate of Disappearance of \(X_2\):**
The pressure decrease due to the disappearance of \(X_2\) is equal to \(P\). Since this change occurred over 10 minutes, we can find the rate of disappearance:
\[ \text{Rate of disappearance of } X_2 = \frac{P}{\text{time}} = \frac{40 \, \text{mm}}{10 \, \text{minutes}} = 4 \, \text{mm/min} \]

### Final Answer:
The rate of disappearance of \(X_2\) is **4 mm per minute**.

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