first order reaction is 25% complete in 40 minutes. Calculate the
value of rate constant. In what time will the reaction be 80%
completed ?
(b) Define order of reaction. Write the condition under which a
bimolecular reaction follows first order kinetics.
Answers
Answer:
128
Explanation:
25% take 40 min
80% take how many time will take
40/25 *80 =128
The value of rate constant is 0.0072 min⁻¹.
The reaction will be 80% completed in 223.58 minutes.
- The first order reaction is 25% complete in 40 minutes
⇒ 75% reactant is left after 40 minutes.
- so we have [R] = 75% of [R]₀ = (75/100)[R]₀ = (3/4)[R]₀
t = 40 min
Using integrated first order reaction, we have:
k = (2.303/t) log {[R]₀/[R]}
k = (2.303/40) log{[R]₀/(3/4)[R]₀}
k = (2.303/40) log(4/3)
k = 0.0576 (log 4 - log 3)
k = 0.0576 × (0.602 - 0.4771)
k = 0.0576 × 0.125
k = 0.0072 min⁻¹
(a) When the reaction is 80% completed, amount of reactants left = 20% of [R]₀ = (20/100) [R]₀ = [R]₀/5
⇒ t = (2.303/k) log {[R]₀/[R]}
t = (2.303/0.0072) log {[R]₀/([R]₀/5)}
t = 319.86 × log 5
t = 319.86 × 0.699
t = 223.58 minutes
(b) The sum of powers of the concentration of the reactants in the rate law expression is called the order of that chemical reaction.
- The order can be 0, 1, 2,3, and even a fraction.
- A bimolecular reaction can follow first order kinetics if one of the reactants is taken in excess so that its concentration does not change.
- Such a reactant will not contribute to the order of reaction. Thus, a bimolecular reaction will be of first order.