Rate = k[A]⁰[B] If we double the concentration of A and B, the rate of reaction becomes:
Correct answer: B. gets doubled
- A. remains the same
- B. gets doubled
- C. gets tripled
- D. gets halved
Explanation
A is raised to the power zero, so changing its concentration has no effect on the rate, while the rate is directly proportional to B. Doubling both concentrations therefore only doubles the rate.
This question appeared on the UHS MDCAT 2026 paper, which you can sit online with every answer explained.
Report an error
The more specific you are, the faster it gets fixed. A source beats an opinion.
Prefer email? [email protected]
About Order of Reaction
Reaction order shows how the rate depends on the concentration of each reactant, as expressed in the rate law. Coverage includes zero, first and second-order reactions, integrated rate equations, half-life, units of rate constants and the distinction between overall order and molecularity.
Practise Reaction Kinetics
576 free Reaction Kinetics MCQs from Chemistry, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Discussion
Stuck on an option, or know a faster way to get there? Ask or explain it here.
No comments yet. Be the first to explain this one.
Exams that ask Chemistry questions like this
Chemistry is on 14 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
More Order of Reaction questions
For 3rd order reaction rate=k[A][B]² when concentration of both reactant is doubled the new rate becomes:
Consider the reaction: NO2(g)+CO(g) ➔ NO(g)+CO2(g). it shows that the reaction is:
In zero order reactions, the rate is independent of:
The rate equation determined experimentally for this reaction: (CH3)3―C―Br + H2O → (CH3)3―C―OH + HBr, is Rate = k[(CH3)3CBr] Hence it is which of the follwing?