In first order reaction, the rate of reactiondepends on 1st power of one reactant. First order reactions are very important reactions as they are very helpful in medical field and they also help us to solve the mystery of our history. You may think why are we talking about history in the post of chemistry? You must have heard about carbon dating technique used to determine the age of fossils by assessing the decay of 14C in them. This decay of 14C follows first order reaction and by using the half life equation, archaeologist determine the age of fossils. Let’s derive the equation for the rate law of first order reaction and then we will compute the equation for half life of first order reaction.
Consider an equation where Reactant ‘R’ gets converted to product ‘P’
R⟶P
Rate = -dR/dt = k [R]1 .........(1)
-d[R]/dt = k [R]
d[R]/ [R] = -k dt .........(2)
Integrating both sides
ln[R] = -k t + I .........(3)
Where I is the constant of integration
At t=0, R= [R]0 the initial concentration
ln [R]0 = -k × 0 + I
ln[R]0 = I
On putting the value of I in equation 3 we get,
ln[R] = -k t + ln[R]0 .........(4)
-k t = ln[R] - ln[R]0
k t = ln[R]0 - ln[R]
k = 1 ln[R]0 ...........(5)
t [R]
or,
k = 2.303 log [R]0 ...........(6)
t [R]
On putting the units in this equation you can find the units of k. It is sec-1for first order reaction. You can calculate the k by using the equation 5 or 6. If you didn’t know the initial concentration but the reactant concentration is given for two different times, then what will you do? Let’s see,
At time t1, [R] = [R]1
By using equation 4:
ln[R]1 = -k t1 + ln[R]0 ...........(7)
At time t2, [R] = [R]2
ln[R]2 = -k t2 + ln[R]0 ...........(8)
on subtracting eq 8 from eq 7:
ln[R]1- ln[R]2 = {-k t1 + ln[R]0}-{-k t2+ ln[R]0}
ln[R]1- ln[R]2 = -k t1 + ln[R]0 + k t2- ln[R]0
ln[R]1- ln[R]2 = k (t2 – t1)
ln [R]1 = k (t2 – t1) ...........(9)
[R]2
 |
| Graphs of 1st order reaction |
Half life of first order reaction
At t = t ½ , [R] = ½ [R]0
On putting the values in equation 6
k = 2.303 log [R]0
t ½ ½ [R]0
t ½ = 2.303log [R]0
k ½ [R]0
t ½ = 2.303log 2[R]0
k [R]0
t ½ = 2.303× 0.301
k
t ½ = 0.693 ...........(10)
k
Half life of first order reaction is independent of reactant concentration. It means that even if we reduce the reactant concentration, the half life of reaction remains constant. Let’s practice some problems to gain a better understanding:
Q. A reacts according to first order kinetics, with k = 5×10-5s-1. If initial concentration is 1M find the initial rate of reaction. What will be the rate after 30min?
Rate law of first order reaction is
Rate = -dR/dt = k [R]1
Rate = 5×10-5 sec-1 [1M]
Rate = 5×10-5 M s-1
= Initial rate is 5×10-5 M s-1.
To calculate the rate after 30min (= 1800s) we need to find the reactant concentration [R]1
ln [R]0 = k t
[R]
Or,
ln [R] = -k t + ln[R]0
On raising both side of equation to the exponent e
eln[R] = e-kt +ln[R]0
[R] = [R]0 e-kt ........(11)
[R] = 1M e-(5×10^-5)(1800)
[R] = 0.91M
So the rate will be:
Rate = 5×10-5 s-1[0.91M]
Rate = 4.5 ×10-5 M s-1
The rate of the reaction after ½ h will be 4.5 ×10-5 M s-1.
Q. Sucrose decomposes to form glucose and fructose according to first order rate law, with half life 3.33h at 25°C. What fraction of a sample of sucrose remains after 9h?
By using equation of half life we can find out the value of k. And then putting the value of k and t in equation 11 we can find the fraction of sucrose remaining [R]/[R]0.
t ½ = 0.693 ...........(10)
k
k = 0.693 / 3.33
k = 0.208 h-1
[R] = [R]0 e-kt ........(11)
[R]/[R]0= e-(0.208)9
[R]/[R]0= e-1.87
[R]/[R]0= 0.154
Fraction remaining is 0.154 or 15.4% sucrose remains after 9h.
In the next post of chemical kinetics we will study second order reaction and its half life in detail.
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