Answer:
a) The rate law is:
rate = k[Acetone][Br₂]⁰[H⁺] = k[Acetone][H⁺]
b) The value of k is:
k = 3.86 × 10⁻³ M⁻¹ · s⁻¹
Explanation:
Acetone (M) Br2 (M) H+ (M) Rate (M/s)
0.30 0.050 0.050 5.7 x 10-5
0.30 0.10 0.050 5.7 x 10-5
0.30 0.050 0.10 1.2 x 10-4
0.40 0.050 0.20 3.1 x 10-4
0.40 0.050 0.050 7.6 x 10-5
A generic rate law for this reaction could be written as follows:
rate = k[Acetone]ᵃ[Br₂]ᵇ[H⁺]ⁿ
The rate for the reaction in trial 2 is:
rate 2 = 5.7 ×10⁻⁵M/s = k(0.3)ᵃ(0.1)ᵇ(0.050)ⁿ
For the reaction in trial 1:
rate 1 = 5.7 ×10⁻⁵M/s = k(0.3)ᵃ(0.050)ᵇ(0.050)ⁿ
If we divide both expressions, we can obtain "b": rate2 / rate1:
rate2/rate1 = k(0.3)ᵃ(0.1)ᵇ(0.050)ⁿ / k(0.3)ᵃ(0.050)ᵇ(0.050)ⁿ
1 = 2ᵇ
b = 0
If we now take the expressions from trial 3 and 1 and divide them, we can obtain "n":
rate 3/rate 1 = k(0.3)ᵃ(0.050)⁰(0.01)ⁿ/ k(0.3)ᵃ(0.050)⁰(0.050)ⁿ
2.1 = 2ⁿ Applying ln to both side of the equation:
ln 2.1 = n ln2
ln2.1/ln2 = n
1 ≅ n
Taking now the reaction in trial 5 and 1 and dividing them:
rate 5/rate 1 = k(0.4)ᵃ(0.050)⁰(0.050) / k(0.3)ᵃ(0.050)⁰(0.050)
4/3 = 4/3ᵃ
a = 1
a)Then the rate law can be written as follows:
rate = k[Acetone][Br₂]⁰[H⁺]
It might be suprising that the rate of bromination of acetone does not depend on the concentration of Br₂. However, looking at the reaction mechanism, you can find out why.
b) Now, we can find the constant k for every experiment and calculate its average value:
rate / [Acetone][Br₂]⁰[H⁺] = k
For reaction 1:
k1 = 5.7 ×10⁻⁵M/s / (0.3 M)(0.050 M) = 3.8 ×10⁻³ M⁻¹ · s⁻¹
Reaction 2: k2 = 5.7 ×10⁻⁵M/s / (0.30 M)(0.050 M) = 3.8 ×10⁻³ M⁻¹ · s⁻¹
Reaction 3: k3 = 1.2 ×10⁻⁴M/s / (0.30 M)(0.10 M) = 4.0 ×10⁻³ M⁻¹ · s⁻¹
Reaction 4: k4 = 3.1 ×10⁻⁴M/s / (0.40 M)(0.20 M) = 3.9 ×10⁻³ M⁻¹ · s⁻¹
Reaction 5: k5 = 7.6 ×10⁻⁵M/s / (0.4 M)(0.05 M) = 3.8 ×10⁻³ M⁻¹ · s⁻¹
Averge value of k:
k = (k1 + k2 + k3 + k4 + k5)/5 = 3.86 × 10⁻³ M⁻¹ · s⁻¹
