Question

A certain reaction has an activation energy of 39.79 kJ/mol. At what Kelvin temperature will the reaction proceed 6.50 times faster than it did at 339 K?

Answer 1

According to Arrhenius, we can relate the rate constants as following:

K = A*exp(-Ea/(RT))

Where:

K = rate constant at Temperature “T”

A = Frequency Factor

E = Activation Energy in J/mol

R = ideal gas constant, 8.314 J/mol-K

T = absolute temperature

Ahrrenius Equation 2 Points

From Ahrrenius equation;

K1 = A*exp(-Ea/(RT1))

K2 = A*exp(-Ea/(RT2))

Note that A and Ea are the same, they do not depend on Temperature ( in the range fo temperature given)

Then

Divide 2 and 1

K2/K1 = A/A*exp(-Ea/(RT2)) / exp(-Ea/(RT1))

Linearize:

ln(K2/K1) = -Ea/R*(1/T2-1/T1)

get rid of negative sign

ln(K2/K1) = Ea/R*(1/T1-1/T2)

substitute values and solve

K2 = 6.5*K1

K1 = K1

R = 8.314 J/molK

Ea = 39.79 kJ/mol = 39790 J/mol

T1 = 339 K, T2 = ?

substitute

ln(K2/K1) = Ea/R*(1/T1-1/T2)

ln(6.5*K1/K1) = 39790/8.314*(1/339-1/T2)

ln(6.5) = 39790/8.314*(1/339-1/T2)

ln(6.5)/39790 * 8.314 - 1/339 = -1/T2

T2 = -(-0.002558^-1) = 390.34 K

T2 = 390.34 K