Question

The reaction A + B → 2C has the rate law rate = k[A][B]³. By what factor does the rate of reaction increase when [A] remains constant but [B] is doubled?

Answer 1

When the concentration of B is doubled while A remains constant in the reaction A + B → 2C with the rate law rate = k[A][B]³, the reaction rate increases by a factor of 8.

Step-by-step explanation:

The rate law for the reaction A + B → 2C is given by rate = k[A][B]³. If the concentration of A remains constant but the concentration of B is doubled, we need to determine how much the reaction rate increases. According to the rate law, the rate is proportional to the third power of [B]. So, if [B] is doubled (multiplied by 2), the new rate will be:

rate' = k[A](2[B])³ = k[A] × 8[B]³ = 8 × (k[A][B]³)

This means that doubling [B] will increase the reaction rate by a factor of 8, assuming [A] is held constant