Question

1) 2A → B + C

Two trials of the above reaction are run. The concentration of A in the second trial is three times that in the first. It is found that the initial rate of the reaction in the second trial is nine times the initial rate in the first trial. This indicates that the reaction is

- zero order in [A]

- first order in [A]

- second order in [A]

- third order in [A]

2)The following reaction is run and the data obtained:

2NO(g) + 2H₂(g) → N₂(g) + 2H₂)(g)

Trial	PNO	PH2	Initial Rate
1	1 atm	1 atm	0.02 atm/s
2	2 atm	1 atm	0.16 atm/s
3	1 atm	2 atm	0.04 atm/s

What is the order of the reaction with respect to H₂?

- 0

-1

-2

-3

3) The following reaction is run and the data obtained:

2NO(g) + 2H₂(g) → N₂(g) + 2H₂)(g)

Trial	PNO	PH2	Initial Rate
1	1 atm	1 atm	0.02 atm/s
2	2 atm	1 atm	0.16 atm/s
3	1 atm	2 atm	0.04 atm/s

What is the order of the reaction with respect to NO?

4) For the reaction A + B → C the following data is obtained:

Trial	[A]	[B]	RateA
1	0.2 M	0.2 M	1.2 x 10-3 M/min
2	0.1 M	0.2 M	6.0 x 10-4 M/min
3	0.2 M	0.1 M	1.2 x 10-3 M/min

After you determine the rate law, what is the value of the rate constant for the reaction using the given units? If you choose to enter your answer in scientific notation you must enter it this way 1.2E-3 for 1.2 x 10^-3.

Answer 1

1) Let,

Rate = K [A]^n

Initial Rate = R

Final Rate = 9R

Initial Conc. = C

Final Conc. = 3C

R = K x C^n

9R = K x (3C)^n

Dividing the above two equations we get,

3^n = 9

=> n = 2

=> Reaction is Second Order in A

2) Let rate of reaction be,

R = K [NO]^x [H2]^y

From trial 1,2 and 3 we get

0.02 = K [1]^x [1]^y ..............(1)

0.16 = K [2]^x [1]^y .............(2)

and 0.04 = K [1]^x [2]^y ............ (3)

Dividing 1 by 2 gives

2^x = 8

=> x = 3

Dividing 1 by 3 gives,

2^y = 2

=> y = 1

Therefore, reaction is 1st order wrt H2

3) Since x = 3, Reaction is 3rd order wrt NO

4) From trials 1,2 and 3

1.2 x 10^-3 = K [0.2]^x [0.2]^y ............... (1)

6 x 10^-4 = K [0.1]^x [0.2]^y ...................(2)

1.2 x 10^-3 = K [0.2]^x [0.1]^y ...............(3)

Dividing 2 by 1 gives,

2^x = 2

=> x = 1

Dividing 3 by 1 gives,

2^y = 1

=> y = 0

Therefore,

Rate = K [A]

1.2 x 10^-3 = K x 0.2

=> K = 6 x10^-3 or 6E-3