Question

Use the following data to answer Questions 25 and 26: A general reaction written as A + 2B → C + 2D is studied and yields the following data: [A]0 [B]0 Initial Δ[C]/Δt 0.150 M 0.150 M 8.00 × 10–3 mol/L·s 0.150 M 0.300 M 1.60 × 10–2 mol/L·s 0.300 M 0.150 M 3.20 × 10–2 mol/L·s 25. What is the numerical value of the rate constant?

Answer 1

Answer – We are given, a general reaction - A + 2B -----> C + 2D

[A]o	[B]o	∆[C]/ ∆t
0.150	0.150	8.00*10-3
0.150	0.300	1.60*10-2
0.300	0.150	3.20*10-2

First we need to calculate the order of the reaction for calculating the rate constant

So assume the rate law

Rate = k [A]^m [B]ⁿ

We assume m and n order with respect to A and B, so rate law are as follow -

Rate1 = k [A]₁^m [B]₁ⁿ

Rate2 = k [A]₂^m [B]₂ⁿ

Rate 3 = k [A]₃^m [B]₃ⁿ

Now first we need to calculate order of A

So,

Rate3/ Rate1 = k [A]₃^m [B]₃ⁿ / k [A]₁^m [B]₁ⁿ

3.20*10^-2 / 8.0*10^-3 = (0.300)^m /(0.150)^m * (0.150)ⁿ / (0.150)ⁿ

3.75 = (2)^m

So, m = 2

So order with respect to A is second order

Now order of B

Rate2/Rate1 = k [A]₂^m [B]₂ⁿ / k [A]₁^m [B]₁ⁿ

1.60*10^-2 / 8.0*10^-3 = (0.150)^m /(0.150)^m *(0.300)ⁿ /(0.150)ⁿ

2 = (2)ⁿ

So, n = 1

So order with respect B is first

So overall order = 2+1 = 3

So rate law, rate = k [A]²[B]

So overall order is third order

Now we need to calculate the rate constant

We know rate law

Rate = k [A]²[B]

8.0*10^-3 M/s = k *(0.150 M)²(0.150)

k = 8.0*10^-3 M.s^-1/ (0.150 M)²(0.150M)

= 2.37 M^-2 s^-1

So, rate constant is 2.37 M^-2 s^-1