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:

Increase by a factor of 4

Given that the reaction is second-order with respect to the concentration of carbon monoxide (CO), the rate of reaction can be represented as:
\[ \text{Rate} = k[CO]^2 \]
Now, let's analyze the effect of doubling the concentration of CO on the rate of reaction:
Doubling the concentration of CO: \( [CO] \rightarrow 2[CO] \)
Substituting this change into the rate expression:
\[ \text{New Rate} = k[(2[CO])]^2 = k(4[CO]^2) = 4k[CO]^2 \]
Comparing the new rate (\( 4k[CO]^2 \)) to the earlier rate (\( k[CO]^2 \)), we can see that the new rate is four times greater than the earlier rate. Therefore, the correct answer is:
(3) Increase by a factor of 4
Doubling the concentration of CO results in a fourfold increase in the rate of the reaction, assuming all other factors remain constant.