1. Assume that NO(g) and Hsub2(g) react according to the rate expression: rate = k[NO]^2[Hsub2]. How does the rate change if the volume of the enclosing vessel is suddently halved?
If the volume of enclosing vessel is reduced, there will be greater molecules per unit volume, thus higher frequency of effective collision, leading to greater rate of reaction.
2. It has been determined that the reaction in the following data is also first order with respect to the acid catalyst. Propose a mechanism for the slow step consistent with the data:
Initial Concentration.
[CHsub3COCHsub3]
.00M
.0500M
.0500M
[Isub2]
.100M
.100M
.500M
Initial Rate.
1.16 * 10^-7
5.79 * 10^-8
5.78 * 10^-8
Any help would be great, I've been extremely confused with rate law expressions this whole unit. Thanks in advance.
Lets say u find out that r = k[A]^2 [ B]
In the slow step, the order of reactants is the stoichiometric coefficient of reactants. (You can use this to check for consistency of proposed mechanism with the rate law)
So you would expect 2 moles of A to reacts with 1 mole of B in the slow step.
Kelvin