Part A The activation energy of a certain reaction is 40.9 kJ/mol . At 30 C , th
ID: 886114 • Letter: P
Question
Part A
The activation energy of a certain reaction is 40.9 kJ/mol . At 30 C , the rate constant is 0.0130s1 . At what temperature in degrees Celsius would this reaction go twice as fast?
Express your answer with the appropriate units.
Part B
Given that the initial rate constant is 0.0130s1 at an initial temperature of 30 C , what would the rate constant be at a temperature of 200 C for the same reaction described in Part A?
Express your answer with the appropriate units.
Part A Consider the following multistep reaction:
Based on this mechanism, determine the rate law for the overall reaction.
Part B
Consider the following multistep reaction:
Based on this mechanism, determine the rate law for the overall reaction.
Please show all work, thank you
A+BAB(slow) A+ABA2B(fast) 2A+BA2B(overall)Explanation / Answer
To calculate the temperature at which reaction go twice as fast use The Arrhenius Equation as follows:
ln(k/k) = - (Ea/R)[(1/T) - (1/T)]
Here,
T = 30 C or 303 K
k = 0.0130 /s.
k2 = 0.0130 /s *2
Ea = 40.9 kJ/mol or 40900 J/mol
R = 8.314 J/mol•K
– [(1/T) - (1/T)] = (R/Ea)ln(k/k)
Now put all values
–(1/T) + (1/303 K) = [(8.314 J/mol•K)/(40900 J/mol)]ln(2)
(1/T) = [(1/303) – 1.41 × 10] K¹ = 3.16× 10³
T= 316.5 K or 43.5 C
Part B
To calculate the rate constant be at a temperature of 200 C use The Arrhenius Equation as follows:
ln(k/k) = - (Ea/R)[(1/T) - (1/T)]
Here,
T = 30 C or 303 K
T2 = 200 C or 473 K
k = 0.0130 /s.
k2 =?
Ea = 40.9 kJ/mol or 40900 J/mol
R = 8.314 J/mol•K
In k-Ink = - (Ea/R)[(1/T) - (1/T)]
In k-In 0.0130 /s. = - (40900 J/mol/8.314 J/mol•K)[(1/473) - (1/303)K]
In k-(-4.34). = - 4919.12 [(2.11*10^-3) - (3.30*10^-3)K]
In k = 5.85 -4.34=1.51
k = e^1.51= 4.52/s