If the reaction is not reversible to a significant extent, then the number of product species is irrelevant to the rate law. We are always assuming that the reaction are not reversible in all integrated rate laws calculation. If the stoichiometric coefficients were not 1:1, then there would be slight changes to the procedure. Suppose that the reaction is A+2B-->Products. At the start of the reaction, the concentrations of A and B would still be [A]_0 and [B]_0, but after some time has passed, the concentrations would be [A]_0 - x and [B]_0 - 2x, as B reacts twice as fast as A. The differential equation to integrate would then be dx/(([A]_0 - x)([B]_0 - 2x), whose solution you could look up in tables. That solution would indicate that the rate of reaction for B is twice as large as the rate for reaction with A.
i was looking forward to the math here.
same lol
same
same 😢
Would the amount of product (P) that is formed be x or 2x?
Jazaakallahu khaiiran.. thank you teacher 👍
what happens if A and B are not 1:1, and what happens if We had more then one product?
If the reaction is not reversible to a significant extent, then the number of product species is irrelevant to the rate law. We are always assuming that the reaction are not reversible in all integrated rate laws calculation. If the stoichiometric coefficients were not 1:1, then there would be slight changes to the procedure. Suppose that the reaction is A+2B-->Products. At the start of the reaction, the concentrations of A and B would still be [A]_0 and [B]_0, but after some time has passed, the concentrations would be [A]_0 - x and [B]_0 - 2x, as B reacts twice as fast as A. The differential equation to integrate would then be dx/(([A]_0 - x)([B]_0 - 2x), whose solution you could look up in tables. That solution would indicate that the rate of reaction for B is twice as large as the rate for reaction with A.
Kyle Brown products are not taken into account