The rate law for a reaction between substance \(P\) and \(Q\) is given by
\(\text{Rate = }k[P]^x[Q]^y\)
If the concentration of \(P\) is doubled and the concentration of \(Q\) is halved, the ratio of the new rate to the earlier rate of reaction will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(2^{x - y}\)
Initial rate: r1 = k[P]x[Q]y
[P] is doubled and [Q] is halved:
r2 = k(2[P])x(½[Q])y = 2x × 2−y × k[P]x[Q]y
r2 / r1 = 2x × 2−y = 2x−y
NCERT (2026-27 edition):
Rate Expression and Rate Constant – “… The equation like (3.4), which relates the rate of a reaction to concentration of reactants is called rate law or rate expression. Thus, rate law is the expression in which reaction rate is given in terms of molar concentration of reactants with each term raised to some power, which may or may not be same as the stoichiometric coefficient of the reacting species in a balanced chemical equation.”