Question

For the reaction A+B+C→D+E, the initial reaction rate was measured for various initial concentrations of reactants. The following data were collected: Trial [A] (M) [B] (M) [C] (M) Initial rate (M/s) 1 0.10 0.10 0.10 3.0×10−5 2 0.10 0.10 0.30 9.0×10−5 3 0.20 0.10 0.10 1.2×10−4 4 0.20 0.20 0.10 1.2×10−4What is the value of the rate constant k for this reaction? When entering compound units, indicate multiplication of units explicitly using a multiplication dot (multiplication dot in the menu). For example, M−1⋅s−1. Express your answer to two significant figures and include the appropriate units. Indicate the multiplication of units explicitly either with a multiplication dot or a dash.

Answer 1

Answer:

k = 0.030 M⁻²· s⁻¹

Explanation:

Trial    [A] (M)        [B] (M)     [C]( M)    Rate (M/s)

1           0.10           0.10         0.10          3.0 x 10-5

2          0.10           0.10         0.30         9.0 x 10-5

3          0.20          0.10         0.10           1.2 x 10-4

4          0.20          0.20        0.10           1.2 x 10-4

The rate law for the reaction can be written as:

rate = k[A]ᵃ[B]ᵇ[C]ⁿ

To obtain k and the exponents, let´s first write the rates for trials 1 and 2:

rate 1 =  3.0 ×10⁻⁵ = k (0.10)ᵃ·(0.10)ᵇ·(0.10)ⁿ

rate 2 = 9.0 ×10⁻⁵ = k (0.10)ᵃ·(0.10)ᵇ·(0.30)ⁿ     (notice that 0.30 = 3·0.10) Then:

rate 2= 9.0 ×10⁻⁵ = k (0.10)ᵃ·(0.10)ᵇ·(3·0.10)ⁿ

Distributing the exponent:

rate 2 = 9.0 ×10⁻⁵ = k (0.10)ᵃ·(0.10)ᵇ·3ⁿ·(0.10)ⁿ

If we divide rate 2 by rate 1:

rate 2 / rate 1 = 9.0 ×10⁻⁵/ 3.0 ×10⁻⁵ = 3

3 = k (0.10)ᵃ·(0.10)ᵇ·3ⁿ·(0.10)ⁿ / k (0.10)ᵃ·(0.10)ᵇ·(0.10)ⁿ

3 = 3ⁿ

n = 1

In the same way, we can proceed with rate 3 and rate 1 to obtain the exponent "a". Notice that (0.20)ᵃ = 2ᵃ·(0.10)ᵃ

rate 3 / rate 1 =   1.2 x 10⁻⁴ / 3.0 x 10⁻⁵ = 4

4 = k·2ᵃ·(0.10)ᵃ·(0.10)ᵇ·(0.10) /  k (0.10)ᵃ·(0.10)ᵇ·(0.10)

4 = 2ᵃ

a = 2

To obtain "b" we use trial 4 and trial and trial 3:

rate 4 / rate 3 = 1.2 x 10⁻⁴ / 1.2 x 10⁻⁴  = 1

1 = k·(0.2)²·2ᵇ·(0.1)ᵇ·(0.1) / k·(0.2)²·(0.1)ᵇ·(0.1)

1 = 2ᵇ

b = 0

The rate law will be:

rate = k[A]²[B]⁰[C]

Solving for k:

k = rate / [A]²[B]⁰[C]

Now, let´s calculate k for each trial. The value of k will be the average value:

k1 = 3.0 ×10⁻⁵ M/s / (0.10 M)²·(0.10 M)⁰·(0.10 M)

k1 = 3.0 ×10⁻⁵ M/s/ 1.0 × 10⁻³ M³ = 0.030 M⁻²· s⁻¹

k2 = 9.0 ×10⁻⁵ M/s / (0.10 M)²·(0.10 M)⁰·(0.30 M) = 0.030 M⁻²· s⁻¹

k3 = 1.2 x 10⁻⁴ M/s / (0.20 M)²·(0.10 M)⁰·(0.10 M) = 0.030 M⁻²· s⁻¹

k4 =  1.2 x 10⁻⁴ M/s / (0.20 M)²·(0.20 M)⁰·(0.10 M) = 0.030 M⁻²· s⁻¹

The value of k is  0.030 M⁻²· s⁻¹