`color{green}("Definition") :` The rate of the reaction is proportional to the first power of the concentration of the reactant color{red}(R).
`=>` For example,
`color{red}(R → P)`
`color{red}(text(Rate) = - (d [R])/(dt) = k [R]` or `color{red}((d [R])/([R]) = - k dt)`
Integrating this equation, we get `color{red}(ln [R] = - k t +I)` ..........(4)
Again, `color{red}(I)` is the constant of integration and its value can be determined easily.
When `color{red}(t = 0, R = [R]_0)`, where `color{red}([R]_0)` is the initial concentration of the reactant.
Therefore, equation (4) can be written as
`color{red}(ln[R]_0 = - k xx 0+I)`
`color{red}(ln [R]_0 = I)`
Substituting the value of `color{red}(I)` in equation (4)
`color{red}(ln[R] = -k t + ln[R]_0)` .........(5)
Rearranging this equation `color{red}(ln \ \ ([R])/([R]_0) = - k t` or `color{red}(k = 1/t ln \ \ ([R]_0)/([R]))` .........(6)
At time `color{red}(t_1)` from equation (5)
`color{red}(text()^(**)ln [R]_1 = - k t_1 + text()^(**)ln [R]_0)` .......(7)
At time `color{red}(t_2)`,
`color{red}(ln [R]_2 = -k t_2 + ln [R]_0)` .............(8)
where, `color{red}([R]_1)` and `color{red}([R]_2)` are the concentrations of the reactants at time `color{red}(t_1)` and `color{red}(t_2)` respectively.
Subtracting (8) from (7)
`color{red}(ln [R]_1 - ln [R]_2 = -k t_1 - ( -k t_2))`
`color{red}(ln \ \ ( [R]_1)/([R]_2) = k (t_2 - t_1))`
`color{red}(k = 1/(t_2-t_1) ln \ \ ([R]_1)/([R]_2))` ..........(9)
Equation (5) can also be written as :
`color{red}(ln \ \ ([R])/([R]_0) = - k t)`
Taking antilog of both sides `color{red}([R] = [R]_0 e^(-k t))` ............(10)
Comparing equation (5) with `color{red}(y = mx + c)`,
If we plot `color{red}(ln [R])` against `color{red}(t)` we get a straight line with slope `color{red}(= –k)` and intercept equal to `color{red}(ln [R]_0)`.
The first order rate equation can also be written in the form
`color{red}(k = (2.303)/t log \ \ ([R]_0)/([R]))` .............(11)
`color{red}(text()^(**)log\ \ ([R]_0)/([R]) = (k t)/(2.303))`
If we plot a graph between `color{red}(log \ \ ([R]_0)/([R]) vs t)`, (Fig.), the slope `color{red}(= k/2.303)`
`color{red}("Examples") :` (i) Hydrogenation of ethene is an example of first order reaction.
`color{red}(C_2H_4 (g) +H_2 (g) → C_2 H_6 (g))`
`color{red}(text(Rate) = k [C_2H_4])`
(ii) All natural and artificial radioactive decay of unstable nuclei take place by first order kinetics.
`color{red}(text()_(88)^(226)Ra → text()_(2)^(4)He +text()_(86)^(222)Rn)`
`color{red}(text(Rate) = k [Ra])`
(iii) Decomposition of `color{red}(N_2O_5)` and `color{red}(N_2O)` are some more examples of first order reactions.
`color{green}("First Order Gas Phase Reaction") :`
Let us consider a typical first order gas phase reaction `color{red}(A(g) → B (g) +C (g))`.
`=>` Let `color{red}(p_1)` be the initial pressure of A and `color{red}(p_t)` the total pressure at time `‘color{red}(t’).`
`=>` Integrated rate equation for such a reaction can be derived as Total pressure `color{red}(p_t = p_A + p_B + p_C)` (pressure units).
`=>` `color{red}(p_A, p_B)` and `color{red}(p_C)` are the partial pressures of `A, B` and `C`, respectively.
`=>` If `x` atm be the decrease in pressure of `A` at time `t` and one mole each of `B` and `C` is being formed, the increase in pressure of `B` and `C` will also be `x` atm each.
`color{red}(tt(( , A(g) , -> ,B(g) , + , C(g) ), ( text{At} t =0 ,p_1atm , ,0 atm , , 0 atm ), ( text{At time } t , p_1 -x atm , , x atm , , x atm )))`
where, `color{red}(p_1)` is the initial pressure at time `color{red}(t = 0).`
`color{red}(p_t = (p_1-x) +x+x = p_1+x)`
`color{red}(x = (p_t - p_1))`
where, `color{red}(p_A = p_1-x = p_1 - (p_t - p_1))`
` color{red}(= 2p_1-p_t)`
`color{red}(k = ((2.303)/t) (log \ \ p_1/p_A))` ...........(12)
`color{red}( = (2.303)/t log \ \ p_1/(2p_1-p_t))`
`color{green}("Definition") :` The rate of the reaction is proportional to the first power of the concentration of the reactant color{red}(R).
`=>` For example,
`color{red}(R → P)`
`color{red}(text(Rate) = - (d [R])/(dt) = k [R]` or `color{red}((d [R])/([R]) = - k dt)`
Integrating this equation, we get `color{red}(ln [R] = - k t +I)` ..........(4)
Again, `color{red}(I)` is the constant of integration and its value can be determined easily.
When `color{red}(t = 0, R = [R]_0)`, where `color{red}([R]_0)` is the initial concentration of the reactant.
Therefore, equation (4) can be written as
`color{red}(ln[R]_0 = - k xx 0+I)`
`color{red}(ln [R]_0 = I)`
Substituting the value of `color{red}(I)` in equation (4)
`color{red}(ln[R] = -k t + ln[R]_0)` .........(5)
Rearranging this equation `color{red}(ln \ \ ([R])/([R]_0) = - k t` or `color{red}(k = 1/t ln \ \ ([R]_0)/([R]))` .........(6)
At time `color{red}(t_1)` from equation (5)
`color{red}(text()^(**)ln [R]_1 = - k t_1 + text()^(**)ln [R]_0)` .......(7)
At time `color{red}(t_2)`,
`color{red}(ln [R]_2 = -k t_2 + ln [R]_0)` .............(8)
where, `color{red}([R]_1)` and `color{red}([R]_2)` are the concentrations of the reactants at time `color{red}(t_1)` and `color{red}(t_2)` respectively.
Subtracting (8) from (7)
`color{red}(ln [R]_1 - ln [R]_2 = -k t_1 - ( -k t_2))`
`color{red}(ln \ \ ( [R]_1)/([R]_2) = k (t_2 - t_1))`
`color{red}(k = 1/(t_2-t_1) ln \ \ ([R]_1)/([R]_2))` ..........(9)
Equation (5) can also be written as :
`color{red}(ln \ \ ([R])/([R]_0) = - k t)`
Taking antilog of both sides `color{red}([R] = [R]_0 e^(-k t))` ............(10)
Comparing equation (5) with `color{red}(y = mx + c)`,
If we plot `color{red}(ln [R])` against `color{red}(t)` we get a straight line with slope `color{red}(= –k)` and intercept equal to `color{red}(ln [R]_0)`.
The first order rate equation can also be written in the form
`color{red}(k = (2.303)/t log \ \ ([R]_0)/([R]))` .............(11)
`color{red}(text()^(**)log\ \ ([R]_0)/([R]) = (k t)/(2.303))`
If we plot a graph between `color{red}(log \ \ ([R]_0)/([R]) vs t)`, (Fig.), the slope `color{red}(= k/2.303)`
`color{red}("Examples") :` (i) Hydrogenation of ethene is an example of first order reaction.
`color{red}(C_2H_4 (g) +H_2 (g) → C_2 H_6 (g))`
`color{red}(text(Rate) = k [C_2H_4])`
(ii) All natural and artificial radioactive decay of unstable nuclei take place by first order kinetics.
`color{red}(text()_(88)^(226)Ra → text()_(2)^(4)He +text()_(86)^(222)Rn)`
`color{red}(text(Rate) = k [Ra])`
(iii) Decomposition of `color{red}(N_2O_5)` and `color{red}(N_2O)` are some more examples of first order reactions.
`color{green}("First Order Gas Phase Reaction") :`
Let us consider a typical first order gas phase reaction `color{red}(A(g) → B (g) +C (g))`.
`=>` Let `color{red}(p_1)` be the initial pressure of A and `color{red}(p_t)` the total pressure at time `‘color{red}(t’).`
`=>` Integrated rate equation for such a reaction can be derived as Total pressure `color{red}(p_t = p_A + p_B + p_C)` (pressure units).
`=>` `color{red}(p_A, p_B)` and `color{red}(p_C)` are the partial pressures of `A, B` and `C`, respectively.
`=>` If `x` atm be the decrease in pressure of `A` at time `t` and one mole each of `B` and `C` is being formed, the increase in pressure of `B` and `C` will also be `x` atm each.
`color{red}(tt(( , A(g) , -> ,B(g) , + , C(g) ), ( text{At} t =0 ,p_1atm , ,0 atm , , 0 atm ), ( text{At time } t , p_1 -x atm , , x atm , , x atm )))`
where, `color{red}(p_1)` is the initial pressure at time `color{red}(t = 0).`
`color{red}(p_t = (p_1-x) +x+x = p_1+x)`
`color{red}(x = (p_t - p_1))`
where, `color{red}(p_A = p_1-x = p_1 - (p_t - p_1))`
` color{red}(= 2p_1-p_t)`
`color{red}(k = ((2.303)/t) (log \ \ p_1/p_A))` ...........(12)
`color{red}( = (2.303)/t log \ \ p_1/(2p_1-p_t))`