In a first order reaction, the sum of components of the concentration of reactants in a rate expression is equal to one. Let us take a substance (reactant) “A” which decomposes into products. Its initial concentration is “A°” mol/dm3. Let after time “t” the amount of substance left behind is “A” mol/dm3.
A → Product
Rate of reaction (ROR) = r = k [A]1
The rate of reaction depends upon the concentration of A.
Integrated Rate Law
A → Product
Rate = r = k [A]1 ———- (1)
Rate = r = – dA / dt ———-(2)
Compare eq (1) and (2)
– dA / dt = k [A]1
dA / dt = – K dt
∫dA / dt = – ∫K dt
At t=0 the initial concentration of A is A° and after time t the concentration of A is A.
A°A∫dA / dt = – 0t∫K dt
|loge A|AA° = -K |t|t0 ∵ ∫dA / dt = loge A
loge A – loge A° = -K (t-0)
loge A – loge A° = -Kt
loge A = loge A° – Kt (or) ln A = ln A° – Kt
Now convert loge to log10
2.303 log10 A = 2.303 log10 A° – Kt
Divided both sides by 2.303 and we get
log10 A = log10 A° – Kt/2.303
This is the integrated rate law of first order kinetics.
Other forms of Rate Law
log10 A = log10A° – Kt/2.303
log10 A° – log10A = Kt/2.303
log10 (A° / A) = Kt/2.303
t = 2.303/K log10 (A° / A)
As we know that
loge A = loge A° – Kt
loge A – loge A° = – Kt
loge (A / A°) = – Kt
A / A° = e-kt
A = A° e-kt
Half-Life of First Order Reaction
Half-life is a time in which the reaction completes 50%. The initial concentration is A° but when t=t1/2 the final concentration becomes half, A°/2.
Initial concentration = A°
at t=t1/2, A = A°/2
t = 2.303/K log10 (A° / A)
By putting the value of “t” and “A” we get,
t1/2 = 2.303/K log10 (A° / A°/2)
t1/2 = 2.303 log10 (2) / K
t1/2 = 2.303 (0.3) / K
t1/2 = 0.693 / K
Lifetime of Reaction (tlf)
The lifetime of the reaction is a time in which 100% reaction is complete.
At t = tlf , A = 0
tlf = 2.303/K log10 (A° / 0)
tlf = ∞
The first-order reaction can never complete 100%.
Graphical Representation of First Order Reaction
Graph: Conc. of Reactant vs time



Graph: ln [A] vs time
![Graph: ln [A] vs time](https://sp-ao.shortpixel.ai/client/to_webp,q_glossy,ret_img,w_376,h_373/https://www.uochemists.com/wp-content/uploads/2022/11/First-Order-Reaction-Equation-Graph-1.jpg)
![Graph: ln [A] vs time](https://sp-ao.shortpixel.ai/client/to_webp,q_glossy,ret_img,w_376,h_373/https://www.uochemists.com/wp-content/uploads/2022/11/First-Order-Reaction-Equation-Graph-1.jpg)
![Graph: ln [A] vs time](https://sp-ao.shortpixel.ai/client/to_webp,q_glossy,ret_img,w_376,h_373/https://www.uochemists.com/wp-content/uploads/2022/11/First-Order-Reaction-Equation-Graph-1.jpg)
Numericals
In a first order reaction, the concentration of the reactant decrease from 800 mol/dm3 to 50 mol/dm3 in 2×104 s. Find the rate constant.
t = 2.303/K log10 (A° / A)
K = 2.303/t log10 (A° / A)
K = 2.303/2×104 log10 (800 / 50)
K = 2.303/2×104 log10 16
K = 2.303/2×104 log10 24
K = 2.303/2×104 4 (log10 2)
K = 2.303/2×104 4 (0.3)
K = 2.303/2×104 4 (03/10)
K = 13.818 / 105
K = 13.8×10-5
K = 1.38×10-4
A first order kinetics is 20% complete in 5 min. Find the time in which it is 60% complete.
Let A° = 100
20% reaction is complete so the remaining concentration is A = 80
t = 5 min
t = 2.303/K log10 (A° / A)
5 = 2.303/K log10 (100 / 80) ———- (1)
A° = 100
60% reaction is complete so the remaining concentration is A = 40
t = ?
t = 2.303/K log10 (100 / 40) ———- (2)
Divide equation (2) by (1) we get
t/5 = log10 10 – log 4 / log10 10 – log 8
t/5 = log10 10 – log 22 / log10 10 – log 23
t/5 = 1 – 2 log 2 / 1 – 3 log 2
t/5 = 1 – 2 (0.3) / 1 – 3 (0.3)
t/5 = 0.4 / 0.1
t = 4 × 5
t = 20 min
In first order reaction, “A” gives a product. The concentration of “A” changes from 0.1 M to 0.025 M in 40 mins. Find the rate of reaction when the concentration of “A” is 0.01 M.
Rate of reaction =?
Rate of reaction = r = K [A]1
First to find the rate constant, to find the K
t = 2.303/K log10 (A° / A)
K = 2.303/40 log10 (0.1 / 0.025)
K = 2.303/40 log10 4
K = 2.303/40 2 log10 2
K = (2.303) (2) (0.3) / 40
K = (2.303) (2) (3) / 400
K = (2.303) (6) / 400
K = 13.818 / 4 × 100
K = 3.454/100
Rate = K [A]1
Rate = 3.454/100 × 0.01 M
Rate = 3.454/100 × 1/100
Rate = 3.454/10000
Rate = 3.4 × 10-4
Important Points to Remember
- The time taken for 75% completion of the first-order reaction is equal to 2 times taken for 50% completion. (t75% = 2 t50%)
- The time taken for 99.9% completion of the first-order reaction is equal to 10 times taken for 50% completion. (t99.9% = 10 t50%)
- The time taken for 25% completion of the first-order reaction is equal to 0.415 times taken for 50% completion. (t25% = 0.415 t50%)
- The time taken for 87.5% completion of the first-order reaction is equal to 3 times taken for 50% completion. (t87.5% = 3 t50%)
Prove that t75% = 2 t50%.
t = 2.303/K log10 (A° / A)
Let the initial concentration (A°) of reactant is 100, when 50% reaction is goes to completion the remaining concentration of reactant is 50.
t50% = 2.303/K log10 (100 / 50) ———- (1)
When the reaction is 75%complete, the remaining concentration is 25.
t75% = 2.303/K log10 (100 / 25) ———- (2)
Divide (2) by (1)
t75% / t50% = (2.303/K log10 (100 / 25)) / (2.303/K log10 (100 / 50))
t75% / t50% = 2 log10 2 / log10 2
t75% / t50% = 2
t75% = 2 t50%
Prove that t99.9% = 10 t50%.
Let the initial concentration (A°) of reactant is 100, when 50% reaction goes to completion the remaining concentration of reactant is 50.
t50% = 0.693 / K ——— (1)
when 99.9% reaction goes to completion, the remaining concentration of the reactant is 0.1.
t99.9% = 2.303/K log10 (100 / 0.1)
t99.9% = 2.303/K log10 (1000 / 1)
t99.9% = 2.303/K log10 103 ———- (2)
Divide (2) by (1)
t99.9% / t50% = (2.303/K log10 103) / (0.693 / K)
t99.9% / t50% = 2.303 × 3 / 0.693
t99.9% / t50% = 6.909 / 0.693
t99.9% / t50% =10
t99.9% = 10 t50%
Examples of First Order Reaction
- Nuclear physics reaction (Radioactive decay)
- Growth and decay of bacteria
- Inversion of cane sugar in an acidic medium
- Acidic hydrolysis of ester (esterification)
- Dissociation of N2O5
- Dissociation of H2O2