Question

1. The decomposition reaction of NOBr(g) to 2 NO(g) and Br2(g) is second order with k = 0.90

M-1∙s-1.

2 NOBr(g) …….> 2 NO(g) + Br2(g)

What is the half-life of this reaction if the initial concentration of NOBr is 0.78 M?

2. The decomposition reaction of NOBr(g) to 2 NO(g) and Br2(g) is second order with k = 0.90

M-1∙s-1.

2 NOBr(g) …….> 2 NO(g) + Br2(g)

How much time would pass before a 4.00 g sample of NOBr held in a 0.350 L flask had

decomposed to the extent that 0.55 g of NOBr remains?

3. The decomposition reaction of NOBr(g) to 2 NO(g) and Br2(g) is second order with k = 0.90M-1∙s-1.

2 NOBr(g) ……> 2 NO(g) + Br2(g)

How much of a 4.00 g sample of NOBr held in a 0.350 L flask would remain after thirteen

seconds?

4. The decomposition reaction of carbon disulfide, CS2, to carbon monosulfide, CS, and sulfur is

first order with k = 4.89x10-7s-1 at 800.˚C.

CS2 …..> CS + S

What is the half-life of this reaction at 800.˚C?

5. The decomposition reaction of carbon disulfide, CS2, to carbon monosulfide, CS, and sulfur

is first order with k = 4.89x10-7s-1 at 800.˚C.

CS2 ……> CS + S

How many days would pass before

Answer 1

1.

For a second order reaction, the integrated rate law is

Where k = rate constant

t = time

concentration of reactant A at time t

initial concentration of reactant A.

Our reaction is second order with respect to [NOBr].

Half life of a reaction is defined as the time when the concentration is half the initial value.

Hence, putting in the integrated rate law for second order reaction

Given that initial concentration of NOBr = 0.78 M

and rate constant k =

The half life of the reaction is

Hence, half life of the reaction is about 1.4 s

2.

Volume of the reaction flask, V = 0.350 L

Amount of NOBr present initially , = 4.00 g

Molar mass of NOBr, M = 109.91 g/mol

Hence, initial concentration of NOBr can be calculated as

Final amount of NOBr = 0.55 g

Hence, final concentration of NOBr can be calculated as

Given that k =

we can use the integrated rate law for second order reaction to determine the time passed t

Hence, the time passed is about 69 seconds.

3.

Given initial amount of NaBr = 4.00 g

Initial concentration of NaBr in the 0.350 L flask is

Given that rate constant k =

and t = 13 s

We have to find

Hence, using the integrated rate law

Now, the final concentration of NOBr is 0.047 M = 0.047 mol/L

Since the volume of the flask = 0.350 L, and molar mass of NOBr = 109.91 g/mol, the mass of NOBr present after 13 seconds is

Hence, the amount of NOBr left after 13 seconds is 1.80 g approximately.

4.

The given reaction is 1st order.

The rate constant of the reaction is given as

The integrated rate law of first order reaction is

Where t = time

Now, the half life is the time taken for the reactant concentration to be reduced to half its initial value.

Hence, putting in the rate equation

hence, for our reaction the half life can be calculated as

Hence, the half life of the reaction is .

note: question number 5 is incomplete.