A reaction has a rate law, rate = k times A times B squared. What is the overall order of the reaction.
I think that the order is 0. Is this correct?
Sorry I had to post the question using the word times but for some reason this site would not let me post the original equation using the brackets.
Jiskha doesn't allow students to post web addresses. But you don't need a web address to show this. You want rate = k[A][B]2.
The reaction is 1st order in A (because the exponent of A is 1, 2nd order in B (because the exponent of B is 2, and 3rd order over all (because the sum of 1 + 2= 3).
oh I understand you have to add them. Thank you.
To determine the overall order of the reaction, we need to sum the exponents of the reactant concentrations in the rate law. In this case, the rate law is given as rate = k * [A] * [B]^2, where [A] represents the concentration of reactant A and [B] represents the concentration of reactant B.
To find the overall order, we add the exponents of [A] and [B]. In this case, the overall order is 1 + 2 = 3. This means that the overall order of the reaction is 3, not 0.
The rate law tells us how the rate of the reaction depends on the concentrations of the reactants. In this case, the rate is proportional to the concentration of A to the power of 1 and the concentration of B to the power of 2.