The kinetics of the following second-order reaction were studied as a function o

Question

The kinetics of the following second-order reaction were studied as a function of temperature:
C2H5Br(aq)+OH(aq)C2H5OH(l)+Br(aq)

Part A: Determine the activation energy for the reaction.

Part B: Determine the frequency factor for the reaction.

Part C: Determine the rate constant at 10 C.

Express your answer using two significant figures.

Part D: If a reaction mixture is 0.145 M in C2H5Br, and 0.250 M in OH, what is the initial rate of the reaction at 90 C?

Express your answer using two significant figures.

Temperature (C) k (L/mols) 25 8.81×105 35 0.000285 45 0.000854 55 0.00239 65 0.00633

Explanation / Answer

part A:

Activation energy is the amount of energy required to initiate a chemical reaction.

The activation energy can be determined from reaction rate constants at different temperatures by the equation

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

where
Ea is the activation energy of the reaction in J/mol
R is the ideal gas constant = 8.3145 J/K·mol
T1 and T2 are absolute temperatures
k1 and k2 are the reaction rate constants at T1 and T2

lets take last two values:

Step 1- Convert °C to K for temperatures

T = °C + 273
T1 = 55 + 273
T1 = 328 K

T2 = 65 + 273
T2 = 338K

Step 2- Find Ea

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)
ln(0.00633/0.00239) = Ea/8.3145 J/K·mol x (1/328 K - 1/338 K)
ln(2.648) = Ea/8.3145 J/K·mol x 9.020 x 10-5 K-1
0.974 = Ea(1.084 x 10-5 mol/J)
Ea = 0.898 x 105 J/mol

PART B:

if you start with this.
k = A exp(-Ea/RT)

and take the ln of both sides
ln(k) = ln(Aexp(-Ea/RT))

and since ln(ab) = ln(a) + ln(b)...
ln(k) = ln(A) + ln(exp(-Ea/RT)

rearranging a bit.
ln(k) = (-Ea/R) x 1/T + ln(A)

ln(A)=ln(k)+(Ea/RT)

taking T=328K

lnA=ln(0.00239)+(0.898 x 105 /(8.314 x328))

lnA=26.893

therefore, A=e26.893

A=4.78 x1011

PART C:

lnk=lnA - (Ea/RT)

temperature at which the rate is to be calculated=10 degree celcius=283 K

we know lnA=26.893

lnK=26.893 - (0.898 x 105 /(8.314 x283)

lnK=-11.273

K=e-11.273

K=1.27 x10-5

PART D:

for this second order reaction,

rate=k[C2H5Br][OH] (as the stoichiometric coefficient is one)

rate constant at 90 degrees:(calculating in same way as part C)

lnk=lnA - (Ea/RT)

temperature at which the rate is to be calculated=90 degree celcius=363 K

we know lnA=26.893

lnK=26.893 - (0.898 x 105 /(8.314 x363)

lnK=-2.861

K=e-2.861

K=0.0572

therefore rate=0.0572 x0.145 x0.250

rate=2.07 x10-3

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