The reaction of pyridine, C5H5N, with methyl chloride, CH3I, to yield C6H8NI produces the following rate data.[C5H5N](M)[CH3I](M)Rate(M/s)1. 00 X 10-41.00 X 10-47.5 X 10-72.00 X 10-42.00 X 10-43.0 X 10-62.00 X 10-44.00 X 10-46.1 X 10-6Write the rate law for this process. Calculate the rate constant. Predict the initial rate for initial concentrations of [C5H5N] = 1.5 X 10-4 M and [CH3I] = 6.1 X 10-5 M

Explanation:

[C5H5N](M)  [CH3I](M)       Rate(M/s)

1.00 X 10-4   1.00 X 10-4    7.5 X 10-7

2.00 X 10-4  2.00 X 10-4    3.0 X 10-6

2.00 X 10-4   4.00 X 10-4   6.1 X 10-6

Doubling the concentration of CH3I causes the rate to increase by a factor of 2. The reaction is first order with respect to CH3I.

Doubling both concentrations increases the rate of reaction by a factor of 4. In addition with the order of CH3I, we can conclude that the reaction is also first order with respect to C5H5N.

Rate Law;

Rate = k [C5H5N] [CH3I]

Rate constant;

Picking experiment 1;

Rate = k [C5H5N] [CH3I]

7.5 X 10-7 = 1.00 X 10-4 * 1.00 X 10-4 * k

k = 7.5 X 10-7 / 1.00 x 10-8

k = 7.5 x 10^1 = 75

Predict the initial rate for initial concentrations of [C5H5N] = 1.5 X 10-4 M and [CH3I] = 6.1 X 10-5 M

Rate = k [C5H5N] [CH3I]

Rate = 75 * 1.5 X 10-4 * 6.1 X 10-5

Rate = 686.25 x 10-9

Rate = 6.8625 x 10-7M/s

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.