For an exothermic reaction, the rate law at 298K is Rate = k[H2][I2].?

Question

Predict the effect of each of the following changes on the initial rate of the reaction and give the reason for your prediction.
a. increasing the volume of the reaction vessel at constant temperature
b. adding a catalyst
c. increasing the temperature

Answer 1

a. down - fewer collisions per second
b. up - activation energy lowered, more molecules able to react
c. up - more molecules have the activation energy.

Answer 2

The gradual step in the mechanism is the speed-determining step. From that, we can say fee = k3[IH2][I] the placement is that we can't leave [IH2] or [I] in the speed regulation considering the fact that they don't look to be between the unique reactants. instead, we would want to locate equivalent expressions for [IH2] and [I]. word that step 2 of the mechanism is a short equilibrium step (said by using the <--->). At equilibrium, the ahead and opposite charges of that step are equivalent. ahead fee = k2[I][H2] opposite fee = ok-2[IH2] k2[I][H2] = ok-2[IH2] so [IH2] = (k2 / ok-2)[I][H2] From the first quick equilibrium step, k1[I2] = ok-a million[I]^2 [I] = sqrt ((k1 / ok-a million) [I2]) substitute the recent expressions for [IH2] and [I] into our first fee regulation. fee = k3[IH2][I] = ok ((k2 / ok-2)[I][H2]) (sqrt ((k1 / ok-a million) [I2])) lump all the ok's into one consistent, K4 fee = k4[I][H2](sqrt [I2]) = k4(sqrt ((k1 / ok-a million) [I2])[H2](sqrt [I2]) = k5[H2][I2] it really is determination E.