The rate constant (K) is related to activation energy (Ea), frequency factor (A) and temperature (T) in Kelvin by the equation
R = molar gas constant
K = A(e^(-Ea/RT))
Taking natural log of both sides
In K = In A - (Ea/RT)
In K = (-Ea/R)(1/T) + In A
Comparing this to the equation of a straight line; y = mx + c
y = In K, slope, m = (-Ea/R), x = (1/T) and intercept, c = In A
a) From the question, m = (-Ea/R) = -1.10 × (10^4) K
(-Ea/R) = -1.10 × (10^4) = -11000
R = 8.314 J/K.mol
Ea = -11000 × 8.314 = 91454 J/mol = 91.454 KJ/mol
b) c = In A = 33.5
A = e^33.5 = (3.54 × (10^14))/s
c) K = A(e^(-Ea/RT))
A = (3.54 × (10^14))/s, Ea = 91454 J/mol, T = 25°C = 298.15 K, R = 8.314 J/K.mol
K = (3.54 × (10^14))(e^(-91454/(8.314×298.15))) = 0.0336/s
QED!
= 18.234 grams