Question

3. Given the initial rate data for the reaction A + B –––> C, determine the rate expression for the reaction. [A], M [B], M Δ [C]/ Δ t (initial) M/s 0.215 0.150 5.81 x 10–4 0.215 0.300 1.16 x 10–3 0.430 0.150 2.32 x 10–3 A) (Δ[C]/Δt) = 1.80 x 10–2 M –1 s –1 [A][B] B) (Δ[C]/Δt) = 3.60 x 10–2 M –1 s –1 [A][B] C) (Δ[C]/Δt) = 1.20 x 10–1 M –2 s –1 [A][B]2 D) (Δ[C]/Δt) = 5.57 x 10–2 M –3 s –1 [A]2 [B]2 E) (Δ[C]/Δt) = 8.37 x 10–2 M –2 s –1 [A]2 [B]

Answer 1

Given data

[A], M    [B], M                 Δ [C]/ Δ t       (initial) M/s

0.215     0.150                    5.81 x 10–4

0.215     0.300                     1.16 x 10–3

0.430     0.150                    2.32 x 10–3

A) (Δ[C]/Δt) = 1.80 x 10–2 M –1 s –1 [A][B]

B) (Δ[C]/Δt) = 3.60 x 10–2 M –1 s –1 [A][B]

C) (Δ[C]/Δt) = 1.20 x 10–1 M –2 s –1 [A][B]2

D) (Δ[C]/Δt) = 5.57 x 10–2 M –3 s –1 [A]2 [B]2

E) (Δ[C]/Δt) = 8.37 x 10–2 M –2 s –1 [A]2 [B]

Solution :- using the concnetrations and the initial rates we can find the order with respect to each reactant

Lets find the order with reactant A using the data from the trial 1 and 3

Rate3 / rate 1 = ([A]3/[A]1)^m

2.32E-3 / 5.81E-4 = (0.430/0.215)^m

4 = 2^m

Log 4 = m* log 2

Log 4 / log 2 = m

2=m

Now lets calculate the order of reactat B using the values from the trial 1 and 2

Rate 2 / rate 1 = ([B]2/[B]1)^n

1.16E-3 / 5.81E-4 = (0.300/0.150)^n

2 = 2^n

Log 2 / log 2 = n

1=n

So the order with B is 1

So the rate law for the reactiuon is

Rate = K [A]^2[B]

So the correct answer is option E) (Δ[C]/Δt) = 8.37 x 10–2 M –2 s –1 [A]2 [B]