For a chemical reaction involving reactants P and Q , the experimentally determined rate law is given as Rate = k[P]^ 1 2 [Q]² . If the concentration of P is increased by a factor of 4 and the concentration of Q is halved, how will the new rate of reaction compare to the initial rate?

Step-by-step solution

The initial rate of the reaction is given by: R₁ = k[P]^ 1 2 [Q]² The new concentrations are [P]' = 4[P] and [Q]' = [Q] 2 . Substituting these into the rate law, the new rate is: R₂ = k(4[P])^ 1 2 ( [Q] 2 )² R₂ = k 2[P]^ 1 2 1 4 [Q]² R₂ = 1 2 k[P]^ 1 2 [Q]² R₂ = 1 2 R₁ The new rate becomes half of the initial rate. Answer: It will become half of the initial rate.