Question

ate data for the reaction: 2 NO (g) + Br2 (g) → 2 NOBr (g) n below Fl. M 0.100 0.200 0.300 0.300 INH,J.M Initial Rate, M/min Experiment 2 4 0.200 0.200 0.150 0.050 0400mm 0.80 2.55 0.850 rate law for this reaction, and determine the value of K. Show reasoning! (8 points)

Answer 1

rate = k [BF₃]^a [NH₃]^b

Considering experiment 3 and 4, keeping the concentration of [BF₃] constant and increasing the concentration of [NH₃] three times, the rate increases by three times.

So,

3 x rate = k [BF₃]^a [3 NH₃]^b

or, 3 x rate = k [BF₃]^a 3^b [NH₃]^b

or, 3 = 3^b

or, 3¹ = 3^b

or, b = 1

Again, Considering experiment 1 and 2, keeping the concentration of [NH3] constant and increasing the concentration of [BF3] two times, the rate increases by two times.

So,

3 x rate = k [3 BF₃]^a [NH₃]^b

or, 3 x rate = k 3^a [BF₃]^a  [NH₃]^b

or, 3 = 3^a

or, 3¹ = 3^a

or, a = 1

So, the overall rate is expressed as

rate = k [BF₃]¹ [NH₃]¹

or, rate = k [BF₃] [NH₃]

Now, substituting the values from experiment 1, we get;

0.4 M/min = k (0.10 M) x (0.20 M)

or, 0.4 M/min = k (0.02 M²)

or, k = (0.4 M min^-1) / (0.02 M²)

or, k = 20 M^-1 min^-1