The bromination of acetone is acid-catalyzed. ch3coch3 + br2 ch3coch2br + h+ + br -the rate of disappearance of bromine was measured for several different concentrations of acetone, bromine, and h+ ions at a certain temperature.[ch3coch3] [br2] [h+] rate of disappearance of br2 (m/s)0.30 0.050 0.050 5.7 10-50.30 0.10 0.050 5.7 10-50.30 0.050 0.10 1.2 10-40.40 0.050 0.20 3.1 10-40.40 0.050 0.050 7.6 10-5(a) what is the rate law for the reaction? (b) determine the rate constant (k).

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                 0.0500.0505.7 x 10-5

0.30                   0.100.0505.7 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         0.0507.6 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

there are no statements provided but here is why:

sulfur belongs to group 16 because it can be found in nature whether it is in a free or combined state.

hope this

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.