Question

1. The rate law for a given reaction is rate = k[NOBr]2. The rate constant is 1.0 x 10-5 1/M∙s, and the initial concentration was 0.100 M. What is the first half-life of this reaction?

A) 0.50 s B) 6.9 x 104 s C) 1.0 x 10-5 s D) 1.0 x 106 s E) None of these

2. The reaction SO2Cl2 SO2 + Cl2 is first order in SO2Cl2. If the concentration of SO2Cl2 after 198 s is 1.47 x 10-2 M and the rate constant is 2.34 x 10-2 s-1, what was the initial concentration of SO2Cl2?

A) 1.40 M B) 1.53 M C) 4.65 M D) 6.33 x 102 M E) 1.47 x 102 M

3. A catalyst

A) Should be included in a rate law B) Increases the number of collisions that go on to product C) Lowers the activation energy of a reaction D) Is consumed in the course of the reaction E) Two of the above are true

Answer 1

Rate = -d(NOBr)/dt= K[NOBr]2

When integrated the equation becomes

1/[NoBr]= 1/[NoBr]o+ Kt

Wheee [NOBr]0= concentration of NOBr at zero time, [NoBr] is concentration of NoBr at any time t and K is the rate constant.

Half life is defied as the time required for the concentration to drop to 50% of the initial reaction.

1/[NoBr]0/2= 1/[NOBr]o+Kt1/2, t1/2 is the half life

1/[NOBr]0= K*t1/2

Hence half life t1/2= 1/[NOBr]0*K= 1/(0.1* 1*10^-5)= 1*10⁶ sec ( D is correct)

2.

For 1^st order reaction, -dSO2Cl2/dt = K[SO2Cl2]

When integrated, the equation becomes

ln[SO2Cl2]= ln[SO2Cl2]o- Kt (1), K is rate constant

[SO2Cl2]o is concentration of SO2Cl2 at zero time and [SO2Cl2] is concentration of SO2Cl2 at any time

Substituting the values given in eq.1

Ln(1.47*10^-2)= ln[SO2Cl2]o-2.34*10^-2*198

[SO2Cl2]0= 1.53 ( B is correct)

3. A Catalyst is a foreign substance which allows the reaction to proceed through an alternative path. Catalytic reaction occus in a series of steps where catalyst may form in one reaction and may become a product in the second reaction. If catalyst is part of the slowest reaction ( the rate determining step), it is also included in the rate equation.

catalysts do not increase the collisions. No of collisions is increased by only temperatrue. but Catalyst ensures that there are more effetive collisions by allowing the reaction to proceed through an alternative path by decreasing the activation energy.

Since it is a foregin substance it does not get consumed. So A and C are correct ( E is the right answer)