Derive an intergrated rate for the first order reaction.

Explanation

(a) Reaction in which the rate of the reaction is proportional to the first power of the concentration the reactant, R is called first order reaction.
For example : `R rarr P`
Rate `= -(d[R])/(d t) = k[R]`
`(d[R])/([R]) = -kdt`
Integrating the above equation, ln [R] = -kt + I
Where I is the constant of integration
When t = 0, `R = [R]_(0)`
Where `[R]_(0)` is the initial concentration of the reactant
Therefore, equation (1) can be written as,
`ln[R]_(0) = k xx 0 + I`
`ln [R]_(0) = I`
Substituting the value of I in equation (1)
`ln [R] = kt + ln [R]_(0)`
Rearranging this equation
`ln.([R])/([R]_(0)) = -kt`
`k = (1)/(t)ln.([R]_(0))/([R])`
`k = (2.303)/(t)log.([R]_(0))/([R])`
(b) According to collision theory
(i) Reactant molecules most collide with sufficient kinetic energy known as activation energy or threshold energy.
(ii) Reactant molecules must collide with proper orientation.
(a) First order integrated rate equation :
`K = (2.303)/(t)log_(10).(a)((a-x))`
Where, K = rate constant
a = initial concentration
a - x = concentration at time (t).
t = time
x = concentration ion summed during time (t).
Derivation : For a reaction
`A rarr "products"`
A = reactant with initial concentration a.
Then, according to first order rate equation.
`(dx)/(d t) = K.d t`
On integration :
`int(dx)/((a-x)) = K int d t`
where, .C. is integration constant.
When `t = 0`
`C = -log_(e)a`.
Putting value of .C. in eq. no. (2)
`-log_(e)(a-x) = Kt - log_(e)a`
Or, `log_(e)a - log_(e)(a-x) = Kt`
Or `K = (2.303)/(t)log_(10).(a)/((a-x))`