Question

derive the rate equation for first order reaction

Answer 1

In this type of reaction, the sum of the powers of concentrations of reactants in rate law is equal to 1, that is the rate of the reaction is proportional to the first power of the concentration of the reactant. Consider the reaction R → P again. Therefore, the rate law for this reaction is,

We know that [R] = -kt + [R]0 3. Taking log of both sides, we get

ln[R] = -kt + ln[R]0 ………………………….(IV)

∴ ln[R]/[R]0 = -kt …………………………….(V)

∴ k = (1/t) ln [R]0 /[R] ………………………(VI)

Now, consider equation 3 again. At time t1 and time t2, the equation II will be [R]1 = -kt1 + [R]0 and [R]2 = -kt2 + [R]0 respectively, where [R]1 and  [R]2 are concentrations of the reactants at time t1 and t2 respectively. Subtracting second equation from first one, we get

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

Now, taking antilog of both sides of equation V, we get [R] = [R]0e-kt

Comparing this equation with equation of a straight line y = mx + c, if we plot ln [R] against t, we get a straight line with slope = -k and intercept = ln[R]0

On removing natural logarithm from equation VI, the first-order reaction can also be written as,

k = 2.303/t log[R]0 /[R] …………..(VII)

An example of a first-order reaction is hydrogenation of ethene.

C2H4 + H2 → C2H6

If we plot a graph of log[R]0 /[R] against t, we get slope = k/2.303