Answer:
k = 0.030 M⁻²· s⁻¹
Step-by-step 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