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

For any rate-law expression of the form $$r = k\,[\text{reactant}]^{n}$$, the exponent $$n$$ is called the order of the reaction with respect to that reactant.

The problem states that the reaction is second order in carbon monoxide, so $$n = 2$$. Hence the rate law involving the concentration of carbon monoxide is

$$r = k \,[CO]^{2}\quad\ldots(1)$$

Let the initial concentration of CO be $$[CO]_1$$ and the corresponding initial rate be $$r_1$$. From (1),

$$r_1 = k \,[CO]_1^{2} \quad\ldots(2)$$

Now the concentration of CO is doubled, i.e. $$[CO]_2 = 2[CO]_1$$. The new rate $$r_2$$ becomes

$$r_2 = k \,(2[CO]_1)^{2} = k \,4[CO]_1^{2} = 4\bigl(k\,[CO]_1^{2}\bigr) \quad\ldots(3)$$

Comparing (3) with (2),

$$r_2 = 4\,r_1$$.

Thus, doubling the concentration of carbon monoxide increases the rate of reaction by a factor of 4.

Option C which is: increase by a factor of 4