A gaseous reaction : `2A(g) + B(g) to 2C(g)`, Show a derease in pressure from 120 mm to 100 mm in 10 minutes. The rate of appearance of C is

Options

A. 2 mm/min
B. 4 mm/min
C. 10 mm/min
D. 12 mm/min

Explanation

To find the rate of appearance of C in the reaction \(2A(g) + B(g) \rightarrow 2C(g)\), given that the pressure decreases from 120 mm to 100 mm in 10 minutes, we can follow these steps:

### Step 1: Calculate the Change in Pressure
The change in pressure (\(\Delta P\)) can be calculated as follows:
\[ \Delta P = P_{\text{initial}} - P_{\text{final}} = 120 \, \text{mm} - 100 \, \text{mm} = 20 \, \text{mm} \]

### Step 2: Calculate the Rate of Change of Pressure
The rate of change of pressure (\(R\)) over the time period can be calculated using the formula:
\[ R = \frac{\Delta P}{\Delta t} \]
Where \(\Delta t\) is the time period (10 minutes):
\[ R = \frac{20 \, \text{mm}}{10 \, \text{min}} = 2 \, \text{mm/min} \]

### Step 3: Relate the Rate of Change of Pressure to the Rate of Appearance of C
From the stoichiometry of the reaction, we know:
- The rate of disappearance of A is \(-\frac{1}{2} \frac{dP_A}{dt}\)
- The rate of disappearance of B is \(-\frac{dP_B}{dt}\)
- The rate of appearance of C is \(\frac{1}{2} \frac{dP_C}{dt}\)

Since the overall rate of the reaction can be expressed in terms of the change in pressure, we can write:
\[ \text{Rate} = -\frac{1}{2} \frac{dP_A}{dt} = -\frac{dP_B}{dt} = \frac{1}{2} \frac{dP_C}{dt} \]

### Step 4: Substitute the Rate into the Equation
We already calculated the overall rate of the reaction as \(2 \, \text{mm/min}\). Thus:
\[ \frac{1}{2} \frac{dP_C}{dt} = 2 \, \text{mm/min} \]

### Step 5: Solve for the Rate of Appearance of C
To find \(\frac{dP_C}{dt}\), we can multiply both sides by 2:
\[ \frac{dP_C}{dt} = 2 \times 2 \, \text{mm/min} = 4 \, \text{mm/min} \]

### Conclusion
The rate of appearance of C is:
\[ \frac{dP_C}{dt} = 4 \, \text{mm/min} \]

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