The complete rate law for the hydrolysis of benzene sulfonyl chloride in presence of fluoride ion has to be derived.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 11, Problem 113QRT

(a)

Interpretation Introduction

Interpretation:

The complete rate law for the hydrolysis of benzene sulfonyl chloride in presence of fluoride ion has to be derived.

(a)

Expert Solution

Explanation of Solution

The rate is affected by the concentration of Benzenesulfonyl chloride (BSC), but it is kept constant in this experiment. So its role in representing the residual rate is not obvious. Therefore considering this a rate law can be written as follows,

rate = 2.4×10−7mol L−1s−1+k[BSC][F−]i

Where i and k are presently unknown.

In order to study the effect of [F−] in the rate of the reaction, residual rate has to be subtracted from the observed rate.

Experiment[F−]×102(mol/L)Initial rate×107(mol L−1 s−1) Adjusted initial rate×107(mol L−1 s−1)102.42.4−2.4=020.55.45.4−2.4=3.031.07.97.9−2.4=5.542.013.913.9−2.4=11.553.020.220.2−2.4=17.864.025.225.2−2.4=22.875.032.0232.0−2.4=29.6

Therefore the adjusted rate law can be written in the following form.

Adjusted rate = k[BSC][F−1]i

The experiments 2 to 3, 3 to 4, and 4 to 6 has to be compared for getting the order of F−. each of this shows a factor two increase in the concentration as shown below.

2 to 3 : 5.53.0 = 1.83 to 4 : 11.55.5 = 2.14 to 6 : 22.811.5 = 2.0

It is clear that the concentration change and the adjusted rate change are almost equal, so the fluoride ion concentration and adjusted rate are proportional. Though it is only an approximate relationship, a graph can be drawn by taking log of the rate on y axis and log[F−] on x axis. The graph will be linear and the slope will give the order of the reaction.

log[F−]log(adjusted initial rate)(mol L−1s−1)−2.3−6.52−2.00−6.26−1.70−5.939−1.52−5.750−1.40−5.642−1.30−5.529

OWLV2 FOR MOORE/STANITSKI'S CHEMISTRY:, Chapter 11, Problem 113QRT

Figure 1

Therefore it is clear that the reaction is first order with respect to fluoride ion in the adjusted rate law.

So the rate can be written s follows,

rate = 2.4 × 10−7 mol L−1s−1 + k[BSC][F−]

(b)

Interpretation Introduction

Interpretation:

The rate constant has to be calculated for the hydrolysis of benzene sulfonyl chloride in presence of fluoride ion.

(b)

Expert Solution

Explanation of Solution

The rate constant can be determined using the adjusted rate law as follows,

k =adjusted rate[BSC][F−]

The rate constant can be calculated for the all the experiments with non-zero fluoride ion can be calculated as follows,

k2 =adjusted rate[BSC][F−] = 3.0×10−7mol L−1s−1(2×10−4 M)(0.5×10−2mol L−1) = 0.3 M−1s−1

k3 = 5.5×10−7mol L−1s−1(2×10−4 M)(1×10−2mol L−1) = 0.3 M−1s−1

k4 =11.5×10−7mol L−1s−1(2×10−4 M)(0.5×10−2mol L−1) = 0.3 M−1s−1

k5 = 17.8×10−7mol L−1s−1(2×10−4 M)(3.0×10−2mol L−1) = 0.3 M−1s−1

k6 = 22.8×10−7mol L−1s−1(2×10−4 M)(4.0×10−2mol L−1) = 0.3 M−1s−1

k7 =29.6×10−7mol L−1s−1(2×10−4 M)(5.0×10−2mol L−1) = 0.3 M−1s−1

Experiment[F−]×102(mol/L)Adjusted initial rate×107(mol L−1 s−1)adjusted rate constant(M−1s−1)20.53.00.331.05.50.342.011.50.353.017.80.364.022.80.375.029.60.3

Therefore the average value of the rate constant is 0.3 L mol-1s-1.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Complete the reaction in the drawing area below by adding the major products to the right-hand side. If there won't be any products, because nothing will happen under these reaction conditions, check the box under the drawing area instead. Note: if the products contain one or more pairs of enantiomers, don't worry about drawing each enantiomer with dash and wedge bonds. Just draw one molecule to represent each pair of enantiomers, using line bonds at the chiral center. + More... ☐ ☐ : ☐ + G 1. NaOMe Click and drag to start drawing a structure. 2. H +

6. Ammonia reacts with nitrogen monoxide and oxygen to form nitrogen and water vapor. If the rate of consumption of NO is 4.5 mollitermin) (a) Find the rate of reaction (b) Find the rate of formations of N; and HO (c) Find the rate of consumption of NH, and O 4NH: 4NO 0:4: +60

34. Give the expected major product of each of the following reactions. Conc. HI a. CH3CH2CH2OH b. (CH3)2CHCH2CH2OH Conc. HBr H Conc. HI C. OH Conc.HCI d. (CH3CH2)3COH

Answer 2