Advertisements
Advertisements
Question
The rate constant of a first order reaction increases from 4 × 10−2 to 8 × 10−2 when the temperature changes from 27°C to 37°C. Calculate the energy of activation (Ea). (log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)
Advertisements
Solution
Given:
k1 = 4 × 10−2
k2 = 8 × 10−2
T1 = 300 K
T2 = 310 K
Solution:
`log(k_2/k_1)=E_a/(2.303R)[(T_2-T_1)/(T_1T_2)]`
`log((8xx10^(-2)|)/(4xx10^(-2)))=E_a/(2.303R)[(T_2-T_1)/(T_1T_2)]`
`0.301= E_a/(2.303xx 8.314JK^(-1)mol^(-1))[(310-300)/(310xx300)]`
`E_a=(0.301 × 2.303 × 8.314 × 93000)/10`
Ea = 53598.5 J
APPEARS IN
RELATED QUESTIONS
The decomposition of A into product has value of k as 4.5 × 103 s−1 at 10°C and energy of activation 60 kJ mol−1. At what temperature would k be 1.5 × 104 s−1?
Write a condition under which a bimolecular reaction is kinetically first order. Give an example of such a reaction. (Given : log2 = 0.3010,log 3 = 0.4771, log5 = 0.6990).
The chemical reaction in which reactants require high amount of activation energy are generally ____________.
Consider figure and mark the correct option.
Which of the following graphs represents exothermic reaction?
(a)
(b)
(c)
Arrhenius equation can be represented graphically as follows:
The (i) intercept and (ii) slope of the graph are:
Explain how and why will the rate of reaction for a given reaction be affected when the temperature at which the reaction was taking place is decreased.
A first-order reaction is 50% complete in 30 minutes at 300 K and in 10 minutes at 320 K. Calculate activation energy (Ea) for the reaction. [R = 8.314 J K−1 mol−1]
[Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021]
What happens to the rate constant k and activation energy Ea as the temperature of a chemical reaction is increased? Justify.
The rate of a reaction quadruples when temperature changes from 27°C to 57°C calculate the energy of activation.
(Given: R = 8. 314 J K−1 mol−1, log 4 = 0.6021)
