Question

GIVEN THE FOLLOWING DATA FOR THIS REACTION:

NH4+(aq) + NO2-(aq) ---> N2(g) + 2H2O(l)

EXPT	NH4+	NO2-	RATE
1	0.010 M	0.020 M	0.020 M/s
2	0.015 M	0.020 M	0.030 M/s
3	0.010 M	0.010 M	0.005 M/s

a. Calculate the order of reaction with respect to NH4+

b. Calculate the order of reaction with respect to NO2-

c. Calculate the rate constant k

d. Determine the overall order of reaction

e. Determine the rate law

Answer 1

NH4+(aq) + NO2-(aq) ---> N2(g) + 2H2O(l)

Rate law

Rate = K[NH4^+]^m[NO2^-]^n

0.02   = K(0.01)^m(0.02)^n ------------------> 1

0.03    = K(0.015)^m(0.02)^n -------------------> 2

0.005   = K (0.01)^m (0.01)^n ---------------> 3

equation 1 divide by equation 2

0.02/0.03 = K(0.01)^m(0.02)^n/ K(0.015)^m(0.02)^n

0.66         = (0.01/0.015)^m

(0.66)^1      = (0.66)^m

m    = 1

equation 1 divide by equation 3

0.02/0.005 = K(0.01)^m(0.02)^n/ K(0.01)^m(0.01)^n

4                  = (0.02/0.01)^n

(2)^2            = (2)^n

n   = 2

Rate = K[NH4^+]^1[NO2^-]^2

a.

the order of reaction with respect to NH4+ is 1

b.

the order of reaction with respect to NO2- is 2

c.

0.02   = K(0.01)^1(0.02)^2

K        = 0.02/(0.01)^1(0.02)^2   = 5000M^-2 s^-1

d.

the overall order of reaction is 1+2 = 3

e.

Rate = K[NH4^+]^1[NO2^-]^2