Question

For the reaction 2A + 2B → 2C + D, if concentration of A is doubled at constant [B] the rate increases by a factor of 4. If the concentration of B is doubled with [A] being constant the rate is doubled. Write the rate law of the reaction.

Answer 1

Rate = R₁ = k[A]^x [B]^y 

When concentration of A = [2A] and [B] = constant,

R₂ = 4R₁ = k[2A]^x [B]^y

When [A] = constant and concentration of

Hence order with respect to A is 2 and with respect to B is 1. 

By rate law, 

Rate = A: [A]² [B]