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The general laws of rate revolve around the reactant’s concentrations and the rate of the chemical reaction. Another rate law form can also be determined that revolves around the reactants’ time and concentration other than the differential rate law. These are known as integrated laws of rate.
Further, it can be used to determine the reactant’s amount or product available after a certain time frame. With the help of calculus, a chemical reaction’s differential rate law can be integrated based on time. It gives an equation that links the reactant’s product/amount available in the mixture of reactions to the reaction time. This procedure could be simple as well as complicated based on the complication of rate law differential.
Rate ∝ [A]x [B]y
here, x and y might be equal or not equal to the reactant’s stoichiometric coefficients. Therefore, the rate of the reaction is equal to k [A]x [B]y,
where k is the rate constant.
∴ -d[R]/dt = k [A]x [B]y
Rate ∝ [R]0
∴ Rate = -d[R]/dt = k[R]0=
k × 1 ∴ Rate = -d[R]/dt = k
∴ d[R] = -kdt
When we integrate both sides, [R] = -kt + I……………….(I)
where I is the integration constant. At t = 0, the reactant’s concentration R = [R]0. Where [R]o is the reactant’s initial concentration. Putting this value in equation I,
[R]o = -k × 0 + I = I
Substituting this value of I in the equation (I), we get
[R] = -kt + [R]o………………(II)
Comparing equation II with the straight-line equation y= mx +c, if we plot [R] against t, we find a straight line with slope = -k and intercept = [R]o
Therefore, on simplifying equation II, we get
k ={[R]o – [R]}/t……………..(III)
Rate = k[NH3]0 = k
R → P. The rate law:
Rate ∝ [R]
Thus taking
-dR/dt=[R]1
And integrating from 0 to t, we get
ln[R] = -kt + ln[R]0………………………….(IV)
[R0 is initial con] ∴ ln[R]/[R]o = -kt …………………………….(V)
∴ k = (1/t) ln [R]0 /[R] ………………………(VI)
Let us write equation VI two times, t1 and t2.
k = (1/t1) ln [R]0 /[R] 1
k = (1/t2) ln [R]0 /[R]2
Further, by subtracting the 2nd equation from the 1st one:
ln [R]1– ln[R]2 = -kt1 – (- kt2 )
∴ ln[R]1 /[R]2 = k (t2 – t1)
∴ k = [1/(t2 – t1)] ln[R]1 /[R]2
Next, when we take both sides’ antilog of equation V, we find [R] = [R]0e-kt
When we compare the equation with a straight line equation y = mx + c,
if we put ln [R] against t, a straight line will be found with slope = -k and intercept = ln (R)0
When we remove natural logarithm from equation VI, we can also write the first-order reaction as:
k = 2.303/t log[R]0 /[R] …………..(VII)
By putting a graph of log[R]0 /[R] against t, we find the slope = k/2.303
intercept = ln[R]0
C2H4 + H2 → C2H6
Therefore the reaction rate for the above is k [C2H4]. Hence, equations III and VII are the equations of rate constants of zero and first-order reactions. We can find rate constants, initial and final concentrations, and the time taken for the reaction to occur using these reactions.
Rate Law
The rate law depicts the value of the reaction rate that depends on the reactant’s molar concentration. Here, each term maintains its respective values of power. And these power values might or might not be equal to the stoichiometric coefficient. Consider a general reaction, aA + bB → cC + dD, where a, b, c, d represent the stoichiometric coefficients of the products and reactants. Therefore, the rate law for the given reaction above,Rate ∝ [A]x [B]y
here, x and y might be equal or not equal to the reactant’s stoichiometric coefficients. Therefore, the rate of the reaction is equal to k [A]x [B]y,
where k is the rate constant.
∴ -d[R]/dt = k [A]x [B]y
Order of Reaction
The sum of the values of reactants’ concentration powers is known as order of reaction. In this scenario, a zero-order reaction implies the reaction rate is free from the reactants’ concentration.Zero-Order Reaction
So we already know that the reaction rate in a zero-order reaction is independent of the reactants’ concentration. Thus, it means the powers’ sum of the concentrations is zero. It can only be zero when all the powers are zero. Consider a reaction, R → P. Therefore, the rate law of this reaction is,Rate ∝ [R]0
∴ Rate = -d[R]/dt = k[R]0=
k × 1 ∴ Rate = -d[R]/dt = k
∴ d[R] = -kdt
When we integrate both sides, [R] = -kt + I……………….(I)
where I is the integration constant. At t = 0, the reactant’s concentration R = [R]0. Where [R]o is the reactant’s initial concentration. Putting this value in equation I,
[R]o = -k × 0 + I = I
Substituting this value of I in the equation (I), we get
[R] = -kt + [R]o………………(II)
Comparing equation II with the straight-line equation y= mx +c, if we plot [R] against t, we find a straight line with slope = -k and intercept = [R]o
Therefore, on simplifying equation II, we get
k ={[R]o – [R]}/t……………..(III)
Zero-Order Reaction Examples
Zero-order reactions are rare, but they do take place under specific conditions. An example of this type of reaction is the ammonia decomposition, 2NH3 → N2 + 3H2Rate = k[NH3]0 = k
First-Order Reaction
The reaction has the powers’ sum of reactants’ concentrations which is 1 in the law of rate, meaning the reaction rate equals the reactant’s 1st concentration power. Take the below-given reaction:R → P. The rate law:
Rate ∝ [R]
Thus taking
-dR/dt=[R]1
And integrating from 0 to t, we get
ln[R] = -kt + ln[R]0………………………….(IV)
[R0 is initial con] ∴ ln[R]/[R]o = -kt …………………………….(V)
∴ k = (1/t) ln [R]0 /[R] ………………………(VI)
Let us write equation VI two times, t1 and t2.
k = (1/t1) ln [R]0 /[R] 1
k = (1/t2) ln [R]0 /[R]2
Further, by subtracting the 2nd equation from the 1st one:
ln [R]1– ln[R]2 = -kt1 – (- kt2 )
∴ ln[R]1 /[R]2 = k (t2 – t1)
∴ k = [1/(t2 – t1)] ln[R]1 /[R]2
Next, when we take both sides’ antilog of equation V, we find [R] = [R]0e-kt
When we compare the equation with a straight line equation y = mx + c,
if we put ln [R] against t, a straight line will be found with slope = -k and intercept = ln (R)0
When we remove natural logarithm from equation VI, we can also write the first-order reaction as:
k = 2.303/t log[R]0 /[R] …………..(VII)
By putting a graph of log[R]0 /[R] against t, we find the slope = k/2.303
intercept = ln[R]0
First Order Reaction Example
An example of a first-order reaction is the hydrogenation of ethene.C2H4 + H2 → C2H6
Therefore the reaction rate for the above is k [C2H4]. Hence, equations III and VII are the equations of rate constants of zero and first-order reactions. We can find rate constants, initial and final concentrations, and the time taken for the reaction to occur using these reactions.
