A thrid order reaction is first order with respect to A and second order with respect to B. How does the rate of reaction change if the concentration of A is tripled and that of B is double?

Question

kainui · Answer

Well, if you originally have: $$rate = k [A]^1[B]^2$$ but now you're saying we triple the concentration of A and double B? You just plug it in... $$rate = k (3[A])^1(2[B])^2= k* 3[A]^1*4[B]^2=12*k [A]^1[B]^2$$
So maybe following through the algebra of that on your own you can see that we just made the reaction increase 12 times the original rate!

anonymous · Answer

Thank you man :)

kainui · Answer

No problem. =)

anonymous · Answer

I have another question man, please help me. Assume a reaction proceeding at 27 degree C has its activation energy lowered by 6kJ/mol. By what factor does its rate change?