Answer:
The rate of the reaction increased by a factor of 1012.32
Explanation:
Applying Arrhenius equation
ln(k₂/k₁) = Ea/R(1/T₁ - 1/T₂)
where;
k₂/k₁ is the ratio of the rates which is the factor
Ea is the activation energy = 274 kJ/mol.
T₁ is the initial temperature = 231⁰C = 504 k
T₂ is the final temperature = 293⁰C = 566 k
R is gas constant = 8.314 J/Kmol
Substituting this values into the equation above;
ln(k₂/k₁) = 274000/8.314(1/504 - 1/566)
ln(k₂/k₁) = 32956.4589 (0.00198-0.00177)
ln(k₂/k₁) = 6.92
k₂/k₁ = exp(6.92)
k₂/k₁ = 1012.32
The rate of the reaction increased by 1012.32
Answer:
By a factor of 2.25.
Explanation:
Using the Arrhenius equations for the given conditions:
k1 = A*(exp^(-Ea/(RT1))
k2 = A*(exp^(-Ea/(RT2))
T1 = 231°C
= 231 + 273.15 K
= 574.15 K
T2 = 293°C
= 293 + 273.15 K
= 566.15 K
Ea = 274 kJ mol^-1
R= 0.008314 kJ/mol.K
Now divide the second by the first:
k2/k1 = exp^(-Ea/R * (1/T2 - (1/T1))
= 0.444
2.25k2 = k1
