A reaction was found to be second order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled, with everything else kept the same, the rate of reaction will:

Options

A. remain unchanged
B. triple
C. increase by a factor of 4
D. double

Explanation

To solve the problem, we need to analyze how the rate of a second-order reaction changes when the concentration of one of the reactants is altered. Here are the steps to arrive at the solution:

### Step-by-Step Solution:

1. **Understand the Order of Reaction**: The reaction is second order with respect to carbon monoxide (CO). This means that the rate of reaction (R) is proportional to the square of the concentration of CO. We can express this mathematically as:
\[ R = k [\text{CO}]^2 \]
where \( k \) is the rate constant.

2. **Initial Concentration**: Let's denote the initial concentration of CO as \([CO]\). Therefore, the initial rate of reaction can be written as:
\[ R = k [CO]^2 \]

3. **Doubling the Concentration**: If the concentration of CO is doubled, the new concentration becomes \(2[CO]\). We need to find the new rate of reaction (\(R'\)) with this new concentration:
\[ R' = k (2[CO])^2 \]

4. **Calculate the New Rate**: Now, we can simplify the expression for \(R'\):
\[ R' = k (2[CO])^2 = k \cdot 4[CO]^2 = 4k [CO]^2 \]

5. **Relate New Rate to Initial Rate**: We can relate \(R'\) to the initial rate \(R\):
\[ R' = 4R \]

6. **Conclusion**: This shows that when the concentration of carbon monoxide is doubled, the rate of reaction increases by a factor of four. Therefore, the correct answer is that the rate of reaction will increase by a factor of four.

### Final Answer:
The rate of reaction will increase by a factor of four.

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