Question

A. The activation energy of a certain reaction is 42.0 kJ/mol . At 29 ∘C , the rate constant is 0.0190s^-1. At what temperature in degrees Celsius would this reaction go twice as fast? Express your answer with the appropriate units.

B. Given that the initial rate constant is 0.0190s^-1 at an initial temperature of 29 C , what would the rate constant be at a temperature of 120.  ∘C for the same reaction described in Part A?

Express your answer with the appropriate units.

Answer 1

Part A

ln(k₂/k₁) = - (Ea/R)[(1/T₂) - (1/T₁)]

T₁ = 273 + 29 = 302 K, so . Ea = 42.0 kJ/mol convert in to moles = 42000 J/mol

R = 8.314 J/mol-K

K2/K1 = 2

Set –[(1/T₂) - (1/T₁)] = (R/Ea)ln(k₂/k₁) and plug in for all the knowns:

–(1/T₂) + (1/302 K) = [(8.314 J/mol•K)/(42000 J/mol)]ln(2)

(1/T₂) = [(1/302) – 2 .88 × 10⁻⁴] K⁻¹

= 3.023 × 10⁻³

T₂= 330.768 K

PartB

here he has given T2 we have to find K2

ln(k₂/k₁) = - (Ea/R)[(1/T₂) - (1/T₁)]

T₁ = 273 + 29 = 302 K, T₂ = 273 + 120 = 393K so k₁ = 0.0190 s-1. Ea = 42.0 kJ/mol convert in to moles = 42000 J/mol

R = 8.314 J/mol-K

substitute all the in the above equation

lnk₂ - ln(0.019) = -[42000 / 8.314] [1/393 - 1/302]

lnk₂ - ln(0.019) = -[5051.72][-0.000766]

lnk₂ +3.9633 = 3.8696

lnk₂ = 3.8696 - 3.9633

lnk₂ = -0.0937

k2 = 0.9105s-1