One more chemistry question!?

Question

Consider the following general equation for a chemical reaction. A(g) + B(g) ----> C(g) + D(g)

1) Describe the two factors that determine whether a collision between molecules of A and B results in a reaction.
2) How would a decrease in temperature affect the rate of the reaction shown above? Explain your answer.
3) Write the rate law expression that would result if the reaction proceeded by the mechanism shown below
A+B<-----> [AB] (fast)
[AB]+B----> C+D (slow)
4) Explain why a catalyst increases the rate of a reaction but does not change the value of the equilibrium constant for that reaction.

Thanks!

Answer 1

1) There are actually four factors: a) the nature of the reactants; b) the concentration of the reactants; c) temperature; and d) the presence of a catalyst.
2) Decrease in temperature would lower the rate of reaction. If you use collision theory, molecules react when they collide with energy high enough to cause a reaction. When you have a lower temperature, molecules aren't moving as quickly, so there are fewer collisions, and when they do collide, there's a smaller chance that the reaction will have enough energy to cause a reaction.
3) The rate law expression would be:
rate = k[A][B]^2
You would substitute A+B for AB in the second equation (which is the rate-determining step) which would give you A + 2B ---> C + D. The rate law expression equals the reactants raised to the power of the stoichiometric coefficient in the rate-determining step.
4) A catalyst changes the pathway for both the forward and reverse reactions, lowering the activation energy. Since it changes the path for both, the reaction reaches equilibrium faster but it does not shift the equilibrium one way or the other.

Answer 2

Catalyst increases the rate of a reaction but does not change the value of the equilibrium constant for that reaction
Reason
Because catayst not only increases rate of forward reaction but also increases rate of backward reaction.And equilibrium constant is ratio of rate constant of forward as well as backward reaction,it remains uneffected.

Answer 3

i took chemistry last year but this must be like college chemistry or something. idk. but u can go to sparknotes they have helps for chemistry