For the reaction `CO(g) +2H_2(g) hArr CH__3OH(g)` . If active mass of CO is kept constant and active mass of `H_2` is tripled, the rate of of forward reaction will become

Options

A. Three times
B. Six times
C. Eight times
D. Nine times

Answer

To solve the problem, we will analyze the effect of changing the concentration of hydrogen gas (H₂) on the rate of the forward reaction of the given chemical equation:

**Reaction:**
\[ \text{CO(g)} + 2\text{H}_2(g) \rightleftharpoons \text{CH}_3\text{OH(g)} \]

### Step-by-step Solution:

1. **Write the Rate Expression:**
The rate of the forward reaction can be expressed using the rate law. For the reaction given, the rate \( r \) can be written as:
\[ r = k [\text{CO}]^1 [\text{H}_2]^2 \]
where \( k \) is the rate constant, \([\text{CO}]\) is the concentration of carbon monoxide, and \([\text{H}_2]\) is the concentration of hydrogen.

2. **Identify Initial Conditions:**
Let the initial concentration of \([\text{H}_2]\) be \( [\text{H}_2]_0 \). Therefore, the initial rate of the reaction can be expressed as:
\[ r_0 = k [\text{CO}] [\text{H}_2]_0^2 \]

3. **Change in Concentration of H₂:**
According to the problem, the concentration of hydrogen is tripled. Thus, the new concentration of hydrogen will be:
\[ [\text{H}_2] = 3[\text{H}_2]_0 \]

4. **Calculate the New Rate:**
Substitute the new concentration of hydrogen into the rate expression:
\[ r_{\text{new}} = k [\text{CO}] [3[\text{H}_2]_0]^2 \]
Simplifying this gives:
\[ r_{\text{new}} = k [\text{CO}] [9[\text{H}_2]_0^2] \]
Thus, we can express the new rate as:
\[ r_{\text{new}} = 9 \cdot k [\text{CO}] [\text{H}_2]_0^2 \]

5. **Relate New Rate to Initial Rate:**
Since \( r_0 = k [\text{CO}] [\text{H}_2]_0^2 \), we can relate the new rate to the initial rate:
\[ r_{\text{new}} = 9 \cdot r_0 \]

### Conclusion:
The rate of the forward reaction will become **9 times** the initial rate when the concentration of \( H_2 \) is tripled.