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**(2) First-order reaction**

Consider the following elementary reaction

A â†’ P

If the reaction is first order with respect to [A], the rate law expression is

k is rate constant

If t = 0, the init ial concentration is [A]0 and the concentration at t = t, is [A], then integrating yields

[A] = [A]_{0} e^{â€“kt}â€¦(i)

or â€¦(ii)

Using this idea, the concentration of product with time for this first-order reaction is :

[P] + [A] = [A]_{0}

[P] = [A]_{0} â€“ [A]

[P] = [A]_{0} â€“ [A]_{0} e^{â€“kt }

[P] = [A]_{0} (1 â€“ e^{â€“kt}) â€¦.(iii)

**Graph representation of first order reaction**

[A] = [A]_{0} e^{â€“kt}

**Plot of concentration vs time.**

ln [A] = â€“kt + ln [A]_{0}

**Plot of log [A] vs time.**

**t _{1/2} i.e. half life t ime of first order reaction**

when t = t_{1/2}; then [A] =

ln 2 = kt_{1/2}

**Problem. The half life for the first order decomposition of N _{2}O_{5} is 2.05 ï¿½ 10^{5 }s. How long will it take for a sample of N_{2}O_{5} to decay to 60% of its initial value?**

**Sol. **We know that,

The time at which the sample has decayed to 60% of its initial value then

T = 1.51 x 10^{4} s**Problem. Find the t _{3/4} i.e. 3/4 life time of first order reaction.**

**Sol.**

Integrated rate law expression is

when t = t_{3/4} than

then

ln 4 = kt_{3/4}

**(3) Second-order reaction: (Type I) **

Consider the following elementary reaction,

If the reaction is second order with respect to [A], the rate law expression is rate =

k is rate constant

â‡’

If t = 0, the init ial concentration is [A]_{0 }and the concentration at t = t, is [A], then integration yields

â€¦(i)

The concentration of product with time for second order reaction

or

then

â€¦(ii)

t_{1/2} i.e. Half-life t ime of second order reaction (type I)

when t = t_{1/2 }then

â‡’

**Second-order reaction (Type II) **

Second order reactions of t ype II invo lves two different reactants A and B, as fo llows

Assuming that the reaction is first order in both A and B, the reaction rate is

If t = 0 then the init ial concentration are [A]_{0} & [B]_{0 }and the concentration at t = t, are [A] & [B].

The loss of reactant i.e. the formation of product is equal to

[A]_{0} â€“ [A] = [B]_{0} â€“ [B] = [P]

[B]_{0} â€“ [A]_{0} + [A] = [B]

then

â‡’

the integration yield

let Î” = [B]_{0} â€“ [A]

The solution to the integral involving [A] is given by

Using this so lut ion to the integral, the integrated rate law expressio n beco mes

â‡’

**Graph representation of second order reaction of type I**

Y = mx + C

**Plot of concentration vs time **

(4) n^{th} order reaction where n â‰¥ 2 :

An nth order reaction may be represented as

the rate law is,

where k is rate constant for nth order reaction

â‡’

If at t = 0, the init ial concentration is [A]_{0} and the concentration at t = t, is [A], then integration yields

Let

â€¦(1)

t_{1/2} i.e. Half life time of nth order reaction

Where t = t_{1/2} then

â€¦(2)

i.e. â€¦(3)

Thus we can say that t_{1/2 }of the reaction is inversely proportional to the init ial concentration of reactant, except first order reaction.

So, for a first order reaction (n = 1), t_{1/2} is independent on [A]_{0} for a second order reaction (n = 2), t_{1/2} is dependent on [A]_{0}

for a nth order reaction

**Note :** For the elementary reaction, the order of reaction is equal to the molecularity of the reaction.

**Problem. Find the rate law for the following reaction.**

**Sol.**

Rate law is

**Problem. Find the rate law for the following reaction.**

**Sol. **

(1)

(2)

(3) = k_{1}[A] + k_{2}[A] = (k_{1} + k_{2})[A]

**Problem. Find the rate law for the following reaction**

**Sol. **

(1)

(2) = k_{1}[A] â€“ k_{2}[B] â€“ k_{3}[B]

= k_{1}[A] â€“ (k_{2} + k_{3}) [B]

(3)

(4)

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