Question

The rate constant of a chemical reaction increased from 0.100 s^-1 to 3.00 s^-1 upon raising the temperature from 25.0 ∘C to 55.0 ∘C.

Part A: Calculate the value of ((1/T2)-(1/T1)) where T1 is the initial temperature and T2 is the final temperature. Express your answer numerically in K^-1

Part B: Calculate the value of ln (k1/k2) where k1 and k2 correspond to the rate constants at the initial and the final temperatures as defined in part A. Express your answer numerically.

Part C: What is the activation energy of the reaction? Express your answer numerically in kilojoules per mole.

Answer 1

Answer:

Step 1: Explanation

The Arrhenius equation allows us to calculate activation energies if the rate constant is known as activation energy term Ea increases, the rate constant k decreases and therefore the rate of reaction decreases

K =Ae-Ea/RT

or,

when two different temperature and two different rate constant is given then we used the following given equation

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

where,

k represents the rate constant,

Ea is the activation energy,

R is the gas constant (8.3145 J/K mol),

and T is the temperature expressed in Kelvin

(A) Step 2: Calculation of temperature

Temperature (T1) =(25+273.15)K = 298.15 K

Temperature (T2) = ( 55+273.15)K = 328.15 K

hence, on substituting the value

=> 1/T2-1/T1

=> (1/328.15 K ) - (1/298.15 K )

=> -3.0663 × 10-4 K-1

(B) Step 3: Calculation of Rate constant

Rate constant (K1) = 0.100 s-1

Rate constant (K2) = 3.00 s-1

hence, on substituting the value

=> ln ( k1/k2)

=> ln ( 0.100 s-1 / 3.00 s-1)

=> -3.4012

(C) Step 5: Calculation of Activation energy

we got

ln ( k1/k2) = -3.4012

1/T2-1/T1 = -3.0663 × 10-4 K-1

R = 8.314 J/mol-1K-1

on substituting the value

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

=> -3.4012 = Ea / 8.314 J/ mol-1 K-1 ( -3.0663 × 10-4 K-1)

=> ( -3.4012 × 8.314 J/ mol-1 K-1 ) /  ( -3.0663 × 10-4 K-1) = Ea

=> Ea = 92220 J/mol = 92.22 kJ/mol

[ note: 1000 J = 1 kJ ]

Hence, the activation energy = 92.22 kJ/mol