Question

In the reaction A + B C, doubling the concentration of A doubles the reaction rate and doubling the concentration of B does not affect the reaction rate. What is the rate law for this reaction?
rate = k[B]
rate = k[A]2
rate = k[A][B]
rate = k[A]

Answer 1

The rate law for the reaction wherein you double the concentration of A doubles the reaction rate and doubling the concentration of B does not affect the reaction rate would be rate = k[A]2. This is because if 2 is multiplied by the concentration, the rate also doubles.

Answer 2

Answer: Option (d) is the correct answer.

Explanation:

Rate law is the rate which is in equilibrium with the concentration of reactants raised to the power of their coefficients.

For example,  $xA + yB \rightarrow zC$

           Rate, r = k $[A]^{x} \times [B]^{y}$    ........ (1)

So, when concentration of A is doubled then rate of reaction also doubles. Hence, then rate law will be as follows.

                  2r = k $[2A]^{x} \times [B]^{y}$     ......... (2)

Dividing equation (1) by (2) we get the following.

      $\frac{r}{2r}$ = $\frac{k[A]^{x} \times [B]^{y}}{k[2A]^{x} \times [B]^{y}}$

Therefore, cancelling common factors we get the following.

               $\frac{1}{2} = \frac{1}{2x}$

                 $2^{x}$ = 2

                             x = 1

Thus, we can conclude that in the reaction A + B C, doubling the concentration of A doubles the reaction rate and doubling the concentration of B does not affect the reaction rate then rate of reaction will be as follows.

                            rate = k[A]