For a zero order reaction, the rate of reaction is independent of reactant concentration. It means that the rate remains constant throughout the reaction. How is it possible that the reactant concentration doesn’t affect the rate of reaction? Let’s try to find out the rate equation for zero order reaction:
R⟶ P
Rate = -dR/dt = k [R]0
-d[R]/dt = k × 1
d[R] = -k dt .........(1)
Integrating both sides
[R] = -k t + I .........(2)
Where I is the constant of integration
At t=0, R= [R]0 the initial concentration
[R]0 = -k × 0 + I
[R]0 = I
On putting the value of I in equation 1 we get,
[R] = -k t + [R]0 .........(3)
-k t = [R] - [R]0
k t = [R]0 - [R]
k = [R]0 - [R] ...........(4)
t
If you fill units in equation 4 you will find the unit of rate constant for zero order reaction, which is mol L-1sec-1. When we plot a graph between reactant concentration and time, we will get a straight line with negative slope (-k). That means after a certain time reactant concentration will become negative. But concentration can never be negative. It means zero order is not an actual order of reaction but it seems to be for a limited period of time. You will understand it better by its example.
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| Graphs of Zero Order Reaction |
Decomposition of ammonia NH3on a hot (1130K) platinum surface (Pt) is an example of zero order reaction.
2NH3(g) ⟶ N2(g) + 3H2(g)
Platinum metal acts as a catalyst for this reaction. Ammonia molecules get attached to the surface of hot platinum metal and get decomposed into N2 and H2. Platinum metal has a limited surface area so the other molecules have to wait for their turn. The rate of reaction depends only on the ammonia molecules attached to the surface of Pt metal. When all the sites of Pt get saturated with NH3 molecules, it doesn’t matter how much ammonia is present in the vessel. If the same reaction is carried out in a different condition, it may have different order. But in this metal catalysed reaction limited availability of metal surface makes rate independent of reactant concentration.
Decomposition of HI on hot gold surface and decomposition of nitrous oxide N2O with catalyst Pt are other examples of zero order reactions. Enzyme catalysed reactions in organisms also have zero order because just like metal surface enzymes also have limited active sites to bind with the reactant. That’s why the rate of reaction doesn’t depend on the reactant concentration.
By using equation 3 you can calculate reactant concentration at any time of the reaction. The time taken for the completion of half of the reaction is known as half life of the reaction. Let’s try to derive equation for the half life of zero order reaction:
At t = t ½ , [R] = ½ [R]0
On putting the values in equation 3
½ [R]0 = -k t ½ + [R]0
-k t ½ = ½ [R]0 - [R]0
k t ½ = [R]0 - ½ [R]0
t ½ = [R]0 /2k ......(5)
Rate of zero order reaction is independent of reactant concentration but its half life depends on initial concentration of reactant.
Let’s try to solve a few problems based on zero order reaction.
Q. For a zero order reaction R⟶P. If the initial concentration of R is 1.5 mol L-1 and after 120 sec the concentration is reduced to 0.75 mol L-1. Determine the rate constant.
[R]0 = 1.5 mol L-1, [R] = 0.75 mol L-1
if you look at the values of [R]0and [R] in this problem, you can see
[R] = ½ [R]0
So, t = t ½ =120sec
Equation for half life of zero order reaction is
t ½ = [R]0 /2k
On putting values in it,
120 sec = 1.5 mol L-1 / 2k
k = 1.5 mol L-1 / 240sec
k = 0.00625 mol L-1 sec-1
rate constant will be 0.00625 mol L-1 sec-1 or 0.00625 M sec-1.
Q. If the initial concentration is reduced to 1 mol L-1. Does this affect the half life of above reaction?
From the equation for half life of zero order reaction
t ½ = [R]0 /2k
t ½ ∝ [R]0
t2 / t1 = [R0]2/ [R0]1
t2 / 120 = 1 / 1.5
t2 = 80 sec
New half life will be reduced to 80 sec.
In the next post of chemical kinetics we will study first order reaction and its half life in detail.
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