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 need to analyze the given chemical reaction and how the rate of the forward reaction changes when the concentration of one of the reactants is altered.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction given is:
\[ \text{CO(g)} + 2\text{H}_2(g) \rightleftharpoons \text{CH}_3\text{OH(g)} \]

2. **Understand the Rate Expression**:
The rate of the forward reaction can be expressed in terms of the concentrations (or active masses) of the reactants. For this reaction, the rate of the forward reaction (\(r\)) can be represented as:
\[ r \propto [\text{CO}]^1 \cdot [\text{H}_2]^2 \]
This means the rate is directly proportional to the concentration of CO raised to the power of 1 and the concentration of H2 raised to the power of 2.

3. **Initial Conditions**:
Let the initial concentration of CO be \([CO]\) and the initial concentration of H2 be \([H_2]\).

4. **Change in Concentration of H2**:
According to the problem, the concentration of H2 is tripled. Therefore, the new concentration of H2 will be:
\[ [H_2]_{\text{new}} = 3[H_2] \]

5. **Calculate the New Rate**:
Substituting the new concentration of H2 into the rate expression, we have:
\[ r_{\text{new}} \propto [CO]^1 \cdot (3[H_2])^2 \]
Expanding this gives:
\[ r_{\text{new}} \propto [CO]^1 \cdot 9[H_2]^2 \]

6. **Relate New Rate to Old Rate**:
The original rate was:
\[ r_{\text{old}} \propto [CO]^1 \cdot [H_2]^2 \]
Therefore, we can relate the new rate to the old rate:
\[ r_{\text{new}} = 9 \cdot r_{\text{old}} \]

7. **Conclusion**:
Hence, when the concentration of H2 is tripled, the rate of the forward reaction becomes 9 times the original rate.

### Final Answer:
The rate of the forward reaction will become **9 times**.

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