Can anyone please help me answer this chemistry question?

Question

NO will react with Cl2 according to the reaction: 2NO _(g) + Cl_{2 (g)} → 2NOCl _(g). The following data were obtained experimentally.                     

 

Initial [NO] (mol/L)	Initial [Cl2] (mol/L)	Initial Rate (mol/L.s)
0.010	0.020	9.00 x 10-3
0.010	0.040	1.80 x 10-2
0.020	0.040	7.20 x 10-2

 

a) Determine the rate law for this equation?                         

b) What is the value of k?

c) What is the overall order of the reaction?

d) What would be the initial rate if the initial concentrations of NO and Cl₂ were 0.050 M and 0.075 M respectively?

e) What would happen to the rate if you doubled the concentrations of both reactants?

Answer 1

To determine the rate law, we must first determine the order with respect to both NO and Cl2. We do this as follows:

Compare 2nd and 3rd rows where [NO] doubles from 0.01 to 0.02 mol/L while [Cl2] remains constant at 0.04.

The rate increases FOUR fold from 1.80x10-2 to 7.20x10-2. This tells us the reaction is SECOND order with respect to NO.

Compare 1st and 2nd rows where [Cl2] doubles from 0.02 to 0.04 while [NO] remains constant at 0.01. The rate also doubles (from 9x10-3 to 1.8x10-2). This tells us the reaction is FIRST order with respect to [Cl2]

a) We can now write the rate law as ...

rate = k[NO]²[Cl₂]

b) To find the value of k, simply use any data from the table. I'll use the data in the first row:

rate = k[NO]²[Cl₂]

9.00x10^-3 M/s = k(0.010 M)²(0.020 M)

k = 9.00x10^-3 M/s / 2x10^-6 M³

k = 4500 M^-2s^-1

c) Overall order of reaction = third order (add the exponents)

d) Initial rate when [NO] = 0.050 M and [Cl₂] = 0.075 M

rate = k[NO]²[Cl₂]

rate = 4500 M^-2s^-1 (0.05)²(0.075)

rate = 0.844 M/s

e) Doubling the concentration of both reactants...

doubling [NO] will increase rate 4 x

doubling [Cl₂] will increase rate 2x

doubling both will increase rate 8x