A first order reaction is 50 % complete in 50 minutes at 300 K and the same reaction is again 50 % complete in 25 minutes at 350 K. Calculate activation energy of the reaction.
(Chemistry, Class 12th)
Answers
A first order reaction is 50 % complete in 50 minutes at 300 K and the same reaction is again 50 % complete in 25 minutes at 350 K. The activation energy of the reaction is 12.10 kJ mol-1.
For the first part,
Time taken to complete 50% of the reaction (t) = 50mins.
Temperature T1 = 300K
According to first order kinetics,
Rate constant(k) = (0.693/t)
where t = Half life period (time taken to complete 50% of the reaction).
So, rate constant for the first part, k1 = (0.693/50) = 0.01386min-1
For the second part,
Time taken to complete 50% of the reaction (t) = 25mins.
Temperature T2 = 350K
According to first order kinetics,
Rate constant(k) = (0.693/t)
where t = Half life period (time taken to complete 50% of the reaction).
So, rate constant for the second part, k2 = (0.693/25) = 0.02772min-1
Relation between two rate constants and activation energy is given by the formula as:
log(k2/k1) = [(Ea/2.303R)(1/T1 - 1/T2)] ...(1)
where k2 (0.02772min-1) and k1 (0.01386min-1) are the rate constants of the second and first part of the reaction respectively,
Ea is the activation energy of the reaction,
R is the universal gas constant = 8.314 JK-1 mol-1,
T1 (300K) and T2 (350K) are the temperatures at which the half-life reactions occur.
∴ Putting these values in equation 1, we have,
log(0.02772/0.01386) = [(Ea/2.303X8.314)(1/300 - 1/350)]
⇒ 0.3010 = (Ea/19.15)(1/2100)
⇒ Ea/40215 = 0.3010
⇒ Ea = 12104.72 J mol-1 = 12.10 KJ mol-1.
So, the activation energy of the reaction is 12.10 KJ mol-1.
Know more
Q) A given reaction has an activation energy of 24.52 kj/mol. at 25°c, the half-life is 4 minutes. at what temperature will the half-life be reduced to 20 seconds?
ANSWER) https://brainly.com/question/9472721
A first order reaction is 50 % complete in 50 minutes at 300 K and the same reaction is again 50 % complete in 25 minutes at 350 K. The activation energy of the reaction is 12.10 kJ mol-1.
For the first part,
Time taken to complete 50% of the reaction (t) = 50mins.
Temperature T1 = 300K
According to first order kinetics,
Rate constant(k) = (0.693/t)
where t = Half life period (time taken to complete 50% of the reaction).
So, rate constant for the first part, k1 = (0.693/50) = 0.01386min-1
For the second part,
Time taken to complete 50% of the reaction (t) = 25mins.
Temperature T2 = 350K
According to first order kinetics,
Rate constant(k) = (0.693/t)
where t = Half life period (time taken to complete 50% of the reaction).
So, rate constant for the second part, k2 = (0.693/25) = 0.02772min-1
Relation between two rate constants and activation energy is given by the formula as:
log(k2/k1) = [(Ea/2.303R)(1/T1 - 1/T2)] ...(1)
where k2 (0.02772min-1) and k1 (0.01386min-1) are the rate constants of the second and first part of the reaction respectively,
Ea is the activation energy of the reaction,
R is the universal gas constant = 8.314 JK-1 mol-1,
T1 (300K) and T2 (350K) are the temperatures at which the half-life reactions occur.
∴ Putting these values in equation 1, we have,
log(0.02772/0.01386) = [(Ea/2.303X8.314)(1/300 - 1/350)]
⇒ 0.3010 = (Ea/19.15)(1/2100)
⇒ Ea/40215 = 0.3010
⇒ Ea = 12104.72 J mol-1 = 12.10 KJ mol-1.
So, the activation energy of the reaction is 12.10 KJ mol-1.
Know more
Q) A given reaction has an activation energy of 24.52 kj/mol. at 25°c, the half-life is 4 minutes. at what temperature will the half-life be reduced to 20 seconds?
ANSWER) https://brainly.com/question/9472721