Question

Determine the rate constant at 10 degrees C? If a reaction mixture is 0.160 M in C2H5Br, and 0.260 M in OH−, what is the initial rate of the reaction at 85 ∘C ? The kinetics of the following second-order reaction were studied as a function of temperature:
C2H5Br(aq)+OH−(aq)→C2H5OH(l)+Br−(aq)

Temperature (∘C)	k (L/mol⋅s)
25	8.81×10−5
35	0.000285
45	0.000854
55	0.00239
65	0.00633

Answer 1

Let us solve the numerical graphically

Temperature (∘C)	k (L/mol⋅s)	temperature (Kelvin)	1/ Temperature	ln K
25	8.81E-05	298.15	0.003354	-9.33704
35	0.000285	308.15	0.003245	-8.16302
45	0.000854	318.15	0.003143	-7.06558
55	0.00239	328.15	0.003047	-6.03646
65	0.00633	338.15	0.002957	-5.06246

We will plot a graph between, lnK and 1/Temperature

It will give a straight line which will be compared to the Arrehenius equation

The equation is:

y = -10770x + 26.78

It is like

lnK = ln A - Ea / RT

slope of equation = -Ea / R = -10770

Ea = 10770 X 8.314 = 89541.78 Joules

A0 = arrehenius constant

ln A = intercept = 26.78

A = 4.27 X 10^11

a) Determine the rate constant at 10 degrees C = 283.15 K

putting values in the equation obtained from graph

ln K = -10770x + 26.78

x = 1/T

ln K = -10770 X (1/T) + 26.78

ln K = -10770 X (1/283.15) + 26.78

ln K = -11.25

K = 1.300 X 10^-5

b) If a reaction mixture is 0.160 M in C2H5Br, and 0.260 M in OH−, what is the initial rate of the reaction at 85 ∘C

85 C = (273.15 + 85) K = 358.15 K

let us calculate the rate constant firs

l n K = -10770 X (1/T) + 26.78 = -10770 X (1/358.15) + 26.78 = -3.29

K = 0.0372

Rate = K [ C2H5Br] [ OH-]

Rate = 0.0372 X 0.160 X 0.260 = 0.00155 M sec-1