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

Answer: Option (d) is the correct answer.

Step-by-step 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) * [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) * [B]^(y)` ......... (2)

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

`(r)/(2r)` =
`(k[A]^(x) * [B]^(y))/(k[2A]^(x) * [B]^(y))`

Therefore, cancelling common factors we get the following.

`(1)/(2) = (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]

Answer 2

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.