A gaseous reaction `X_2(g)rarrZ(g)+1/2Y(g)` shows increase in pressure from 150 mm to 170 mm in 10 minutes. The rate of disappearance of `X_2` is

Options

A. `4 "mm min"^(-1)`
B. `16 "mm min"^(-1)`
C. `8 "mm min"^(-1)`
D. `2 "mm min"^(-1)`

Answer

To solve the problem, we need to find the rate of disappearance of \(X_2\) given the change in pressure during the reaction.

### Step-by-Step Solution:

1. **Identify Initial and Final Pressures:**
- Initial pressure \(P_1 = 150 \, \text{mm}\)
- Final pressure \(P_2 = 170 \, \text{mm}\)

2. **Calculate the Change in Pressure:**
\[ \Delta P = P_2 - P_1 = 170 \, \text{mm} - 150 \, \text{mm} = 20 \, \text{mm} \]

3. **Determine the Change in Pressure Due to Reaction:**
- The reaction is \(X_2(g) \rightarrow Z(g) + \frac{1}{2}Y(g)\).
- For every 1 mole of \(X_2\) that reacts, the total pressure increases by \( \frac{1}{2} \) mole of \(Y\) and 1 mole of \(Z\). Therefore, the change in pressure due to the consumption of \(X_2\) can be represented as:
\[ \Delta P = -P + P + \frac{1}{2}P = \frac{1}{2}P \]

4. **Relate Change in Pressure to the Reaction:**
- Let \(P\) be the pressure change corresponding to the amount of \(X_2\) that has reacted. The total pressure after 10 minutes is given as 170 mm.
- Thus, we can set up the equation:
\[ 150 + \frac{1}{2}P = 170 \]

5. **Solve for \(P\):**
\[ \frac{1}{2}P = 170 - 150 = 20 \, \text{mm} \]
\[ P = 20 \times 2 = 40 \, \text{mm} \]

6. **Calculate the Rate of Disappearance of \(X_2\):**
- The rate of disappearance of \(X_2\) is defined as the change in pressure over the time interval. The time interval given is 10 minutes.
\[ \text{Rate of disappearance of } X_2 = \frac{\Delta P}{\Delta t} = \frac{40 \, \text{mm}}{10 \, \text{minutes}} = 4 \, \text{mm/min} \]

### Final Answer:
The rate of disappearance of \(X_2\) is \(4 \, \text{mm/min}\).

---