● Integrated Rate Equations

● Zero Order Reaction

● First Order Reaction

● Half-life Period of a Reaction

● Pseudo First Order Reaction

● Zero Order Reaction

● First Order Reaction

● Half-life Period of a Reaction

● Pseudo First Order Reaction

`=>` We know that the concentration dependence of rate is called differential rate equation.

`=>` It is not always convenient to determine the instantaneous rate, as it is measured by determination of slope of the tangent at point ‘t’ in concentration vs time plot. This makes it difficult to determine the rate law and hence the order of the reaction.

`=>` So, for avoiding this difficulty, we integrate the differential rate equation to give a relation between directly measured experimental data, i.e., concentrations at different times and rate constant.

`=>` The integrated rate equations are different for the reactions of different reaction orders. We shall determine these equations only for zero and first order chemical reactions.

`=>` It is not always convenient to determine the instantaneous rate, as it is measured by determination of slope of the tangent at point ‘t’ in concentration vs time plot. This makes it difficult to determine the rate law and hence the order of the reaction.

`=>` So, for avoiding this difficulty, we integrate the differential rate equation to give a relation between directly measured experimental data, i.e., concentrations at different times and rate constant.

`=>` The integrated rate equations are different for the reactions of different reaction orders. We shall determine these equations only for zero and first order chemical reactions.

`color{green}("Definition") :` Zero order reaction means that the rate of the reaction is proportional to zero power of the concentration of reactants.

Consider the reaction :

`color{red}(R → P)`

`color{red}(text(Rate) = (- d [ R])/(dt) = k [R]^0)`

As any quantity raised to power zero is unity

`color{red}(text(Rate) = - ( d [R])/(dt) = k xx 1)`

`color{red}(d [R] = - k dt)`

Integrating both sides

`color{red}([R] = - k t +I)` ...............(1)

where, `color{red}(I)` is the constant of integration.

At `color{red}(t = 0),` the concentration of the reactant `color{red}(R = [R]_0)`, where `color{red}([R]_0)` is initial concentration of the reactant. Substituting in equation (1)

`color{red}([R]_0 = - k xx 0 + I)`

`color{red}([R]_0 = I)`

Substituting the value of `color{red}(I)` in in the equation (1)

`color{red}([R] = - k t + [R]_0)` ............(2)

Comparing (2) with equation of a straight line, `color{red}(y = mx + c)`,

if we plot `color{red}([R])` against `color{red}(t)`, we get a straight line (Fig.) with slope `color{red}(= –k)` and intercept equal to `color{red}([R]_0)`.

Further simplifying equation (2), we get the rate constant, `color{red}(k)` as

`color{red}(k = ( [R]_0 - [R])/t)` .................(3)

`=>` Zero order reactions are relatively uncommon but they occur under special conditions.

`=>` Some enzyme catalysed reactions and reactions which occur on metal surfaces are a few examples of zero order reactions.

`=>` The decomposition of gaseous ammonia on a hot platinum surface is a zero order reaction at high pressure

`color{red}(2NH_3 (g) underset (Pt catalyst ) overset(1130 K) → N_2 (g) +3H_2 (g))`.

`color{red}(text(Rate) = k [NH_3]^0 = k)`

● In this reaction, platinum metal acts as a catalyst.

● At high pressure, the metal surface gets saturated with gas molecules. So, a further change in reaction conditions is unable to alter the amount of ammonia on the surface of the catalyst making rate of the reaction independent of its concentration.

`=>` The thermal decomposition of `color{red}(HI)` on gold surface is another example of zero order reaction.

Consider the reaction :

`color{red}(R → P)`

`color{red}(text(Rate) = (- d [ R])/(dt) = k [R]^0)`

As any quantity raised to power zero is unity

`color{red}(text(Rate) = - ( d [R])/(dt) = k xx 1)`

`color{red}(d [R] = - k dt)`

Integrating both sides

`color{red}([R] = - k t +I)` ...............(1)

where, `color{red}(I)` is the constant of integration.

At `color{red}(t = 0),` the concentration of the reactant `color{red}(R = [R]_0)`, where `color{red}([R]_0)` is initial concentration of the reactant. Substituting in equation (1)

`color{red}([R]_0 = - k xx 0 + I)`

`color{red}([R]_0 = I)`

Substituting the value of `color{red}(I)` in in the equation (1)

`color{red}([R] = - k t + [R]_0)` ............(2)

Comparing (2) with equation of a straight line, `color{red}(y = mx + c)`,

if we plot `color{red}([R])` against `color{red}(t)`, we get a straight line (Fig.) with slope `color{red}(= –k)` and intercept equal to `color{red}([R]_0)`.

Further simplifying equation (2), we get the rate constant, `color{red}(k)` as

`color{red}(k = ( [R]_0 - [R])/t)` .................(3)

`=>` Zero order reactions are relatively uncommon but they occur under special conditions.

`=>` Some enzyme catalysed reactions and reactions which occur on metal surfaces are a few examples of zero order reactions.

`=>` The decomposition of gaseous ammonia on a hot platinum surface is a zero order reaction at high pressure

`color{red}(2NH_3 (g) underset (Pt catalyst ) overset(1130 K) → N_2 (g) +3H_2 (g))`.

`color{red}(text(Rate) = k [NH_3]^0 = k)`

● In this reaction, platinum metal acts as a catalyst.

● At high pressure, the metal surface gets saturated with gas molecules. So, a further change in reaction conditions is unable to alter the amount of ammonia on the surface of the catalyst making rate of the reaction independent of its concentration.

`=>` The thermal decomposition of `color{red}(HI)` on gold surface is another example of zero order reaction.

`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))`

`=>` 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))`

Q 2937680582

The initial concentration of `N_2O_5` in the following first order reaction `N_2O_5(g) → 2 NO_2(g) + 1/2O_2 (g)` was `1.24 × 10^(–2) mol L^(–1)` at `318 K`. The concentration of `N_2O_5` after 60 minutes was `0.20 × 10^(–2) mol L^(–1)`. Calculate the rate constant of the reaction at `318 K`.

For a first order reaction

`log \ \ ([R]_1)/([R]_2) = (k (t_2-t_1))/(2.303)`

`k = (2.303)/((t_2-t_2)) log \ \ ([R]_1)/([R]_2)`

` = (2.303)/(60 mi n - 0 mi n) log \ \ (1.24 xx 10^(-2) mol L^(-1))/(0.20 xx 10^(-2) mol L^(-1))`

` = (2.303)/(60) log 62. mi n^(-1)`

`k = 0.0304 mi n^(-1)`

Q 2967080885

The following data were obtained during the first order thermal decomposition of `N_2O_5 (g)` at constant volume: `2N_2O_5 (g) → 2N_2 O_4 (g) +O_2 (g)`

Calculate the rate constant.

S. NO. | Time/s | Total Pressure / (atm) |
---|---|---|

1 | 0 | 0.5 |

2 | 100 | 0.512 |

Calculate the rate constant.

Let the pressure of `N_2O_5 (g)` decrease by `2x` atm. As two moles of `N_2O_5` decompose to give two moles of `N_2O_4(g)` and one mole of `O_2(g)` the pressure of `N_2O_4(g)` increases by `2x` atm and that of `O_2 (g)` increases by `x` atm.

`2N_2O_5 (g)` | `→ 2N_2O_4 (g)` | `+ O_2 (g)` | |
---|---|---|---|

Start t = 0 | 0.5 atm | 0 atm | 0 atm |

At time t | (0.5 - 2x) atm | 2x atm | x atm |

`p_t = P_(N_2O_5) +P_(N_2O_4) +P_(O_2)`

` = ( 0.5 - 2x) +2x+x = 0.5+x`

`x = p_t - 0.5`

`P_(N_2O_5) = 0.5 - 2x`

` = 0.5 - 2(p_t - 0.5) = 1.5 - 2P_t`

At `t = 100s ; P_t = 0.512` atm

`P_(N_2O_5) = 1.5-2xx0.512 = 0.476` atm

Using equation (4.16)

`k = (2.303)/(100 s) xx 0.0216 = 4.98 xx 10^(-4) s^(-1)`

`color{green}("Definition") :` The half-life of a reaction is the time in which the concentration of a reactant is reduced to one half of its initial concentration. It is represented as `color{red}(t_(1/2))`.

`color{green}("For the Zero Order Reaction") :` Rate constant is given by equation3.

`color{red}(k = ([R]_0 - [R])/t)`

At `color{red}(t = t_(1/2), [R] = 1/2 [R]_0)`

The rate constant at `color{red}(t_(1/2))` becomes `color{red}(k = ([R]_0 - 1/2 [R]_0)/(t_(1/2)))`

`color{red}(t_(1/2) = ([R]_0)/(2k))`

It is clear that `color{red}(t_(1/2))` for a zero order reaction is directly proportional to the initial concentration of the reactants and inversely proportional to the rate constant.

`color{green}("For the First Order Reaction") :`

`color{red}(k = (2.303 )/t log \ \ ([R]_0)/([R]))`

at `color{red}(t_(1/2) \ \ [R] = [R]_0/2)`

So, the above equation becomes `color{red}(k = (2.303)/t_(1/2) log \ \ ([R]_0)/([R]/2))` or `color{red}(t_(1/2) = (2.303)/k log2))`

`color{red}(t_(1/2) = (2.303)/k xx 0.301)`

`color{red}(t_(1/2) = (0.693)/k)` ..........(13)

`=>` For a first order reaction, half-life period is constant, i.e., it is independent of initial concentration of the reacting species.

`=>` The half-life of a first order equation is readily calculated from the rate constant and vice versa.

● For zero order reaction `color{red}(t_(1/2) prop [R]_0)`, first order reaction `color{red}(t_(1/2))` is independent of `color{red}([R]_0)`.

Table summarises the mathematical features of integrated laws of zero and first order reactions.

`color{green}("For the Zero Order Reaction") :` Rate constant is given by equation3.

`color{red}(k = ([R]_0 - [R])/t)`

At `color{red}(t = t_(1/2), [R] = 1/2 [R]_0)`

The rate constant at `color{red}(t_(1/2))` becomes `color{red}(k = ([R]_0 - 1/2 [R]_0)/(t_(1/2)))`

`color{red}(t_(1/2) = ([R]_0)/(2k))`

It is clear that `color{red}(t_(1/2))` for a zero order reaction is directly proportional to the initial concentration of the reactants and inversely proportional to the rate constant.

`color{green}("For the First Order Reaction") :`

`color{red}(k = (2.303 )/t log \ \ ([R]_0)/([R]))`

at `color{red}(t_(1/2) \ \ [R] = [R]_0/2)`

So, the above equation becomes `color{red}(k = (2.303)/t_(1/2) log \ \ ([R]_0)/([R]/2))` or `color{red}(t_(1/2) = (2.303)/k log2))`

`color{red}(t_(1/2) = (2.303)/k xx 0.301)`

`color{red}(t_(1/2) = (0.693)/k)` ..........(13)

`=>` For a first order reaction, half-life period is constant, i.e., it is independent of initial concentration of the reacting species.

`=>` The half-life of a first order equation is readily calculated from the rate constant and vice versa.

● For zero order reaction `color{red}(t_(1/2) prop [R]_0)`, first order reaction `color{red}(t_(1/2))` is independent of `color{red}([R]_0)`.

Table summarises the mathematical features of integrated laws of zero and first order reactions.

Q 2937180982

A first order reaction is found to have a rate constant, `k = 5.5 × 10^(-14) s^(-1)`. Find the half-life of the reaction.

Half-life for a first order reaction is `t_(1/2) = (0.693)/k`

`t_(1/2) = (0.693)/(5.5xx10^(-14) s^(-1)) = 1.26xx10^(14) s`

Q 2937191082

Show that in a first order reaction, time required for completion of `99.9%` is `10` times of half-life `(t_1/2)` of the reaction.

When reaction is completed `99.9%, [R]_n = [R]_0 – 0.999[R]_0`

When reaction is completed `99.9%, [R]_n = [R]_0 – 0.999[R]_0`

`k = (2.303)/t log \ \ ([R]_0)/([R])`

` = (2.303)/t log \ \ ([R]_0)/([R]_0 - 0.999 [R]_0) = (2.303)/t \ \ log 10^3`

`t = 6.909 // k`

For half-life of the reaction

`t_(1/2) = 0.693 // k`

`t/t_(1/2) = (6.909)/k xx k/(0.693) = 10`

`=>` The order of a reaction is sometimes altered by conditions.

`=>` Consider a chemical reaction between two substances when one reactant is present in large excess.

`color{red}("Examples") :`

(i) During the hydrolysis of `0.01 mol` of ethyl acetate with `10 mol` of water, amounts of the various constituents at the beginning `(t = 0)` and completion `(t)` of the reaction are given as under :

`color{red}(tt(( , CH_3COOC_2H_5 ,+, H_2O , overset(H^+)-> ,CH_3COOH , + , C_2H_5OH ), ( t =0 , 0.01 mol , ,10 mol , ,0 mol , , 0 mol ), ( t , 0 mol , , 9.9 mol , , 0.01 mol , , 0.01 mol )))`

`=>` The concentration of water does not get altered much during the course of the reaction. So, in the rate equation

`color{red}(text(Rate) = k' [CH_3 COOC_2H_5] [H_2O])`

The term `color{red}([H_2O])` can be taken as constant.

Thus, the equation becomes

`color{red}(text(Rate) = k [CH_3COOC_2H_5])`

where `color{red}(k = k' [H_2O])`

and the reaction behaves as first order reaction. Such reactions are called pseudo first order reactions.

(ii) Inversion of cane sugar is another pseudo first order reaction.

`color{red}(undersettext(Cane sugar) (C_(12) H_(22) O_(11) ) + H_2O overset(H^+) →undersettext(Glucose) (C_6H_(12) O_6) + undersettext(Fructose) (C_6H_(12) O_6))`

`color{red}(text(Rate) = k [C_(12) H_(22) O_(11)])`

`=>` Consider a chemical reaction between two substances when one reactant is present in large excess.

`color{red}("Examples") :`

(i) During the hydrolysis of `0.01 mol` of ethyl acetate with `10 mol` of water, amounts of the various constituents at the beginning `(t = 0)` and completion `(t)` of the reaction are given as under :

`color{red}(tt(( , CH_3COOC_2H_5 ,+, H_2O , overset(H^+)-> ,CH_3COOH , + , C_2H_5OH ), ( t =0 , 0.01 mol , ,10 mol , ,0 mol , , 0 mol ), ( t , 0 mol , , 9.9 mol , , 0.01 mol , , 0.01 mol )))`

`=>` The concentration of water does not get altered much during the course of the reaction. So, in the rate equation

`color{red}(text(Rate) = k' [CH_3 COOC_2H_5] [H_2O])`

The term `color{red}([H_2O])` can be taken as constant.

Thus, the equation becomes

`color{red}(text(Rate) = k [CH_3COOC_2H_5])`

where `color{red}(k = k' [H_2O])`

and the reaction behaves as first order reaction. Such reactions are called pseudo first order reactions.

(ii) Inversion of cane sugar is another pseudo first order reaction.

`color{red}(undersettext(Cane sugar) (C_(12) H_(22) O_(11) ) + H_2O overset(H^+) →undersettext(Glucose) (C_6H_(12) O_6) + undersettext(Fructose) (C_6H_(12) O_6))`

`color{red}(text(Rate) = k [C_(12) H_(22) O_(11)])`

Q 2927391281

Hydrolysis of methyl acetate in aqueous solution has been studied by titrating the liberated acetic acid against sodium hydroxide. The concentration of the ester at different times is given below.

Show that it follows a pseudo first order reaction, as the concentration of water remains nearly constant `(55 mol L^(–1))`, during the course of the reaction. What is the value of `k′` in this equation? Rate `= k′ [CH_3COOCH_3][H_2O]`

t/min | 0 | 30 | 60 | 90 |
---|---|---|---|---|

`C//mol L^(-1)` | 0.8500 | 0.8004 | 0.7538 | 0.7096 |

Show that it follows a pseudo first order reaction, as the concentration of water remains nearly constant `(55 mol L^(–1))`, during the course of the reaction. What is the value of `k′` in this equation? Rate `= k′ [CH_3COOCH_3][H_2O]`

For pseudo first order reaction, the reaction should be first order with respect to ester when `[H_2O]` is constant. The rate constant `k` for pseudo first order reaction is

`k = (2.303)/t log \ \ C_0/C`

where `k = k' [H_2O]`

From the above data we note

t/min | `C// mol L^(-1)` | `k//mol L^(-1)` |
---|---|---|

`0` | `0.8500` | `-` |

`30` | `0.8004` | `2.004 xx 10^(-3)` |

`60` | `0.7538` | `2.002xx10^(-3)` |

`90` | `0.7096` | `2.005xx10^(-3)` |

It can be seen that `kappa [H_2O]` is constant and equal to `2.004 × 10^(-3) mi n^(–1)` and hence, it is pseudo first order reaction. We can now determine `k` from

`k [H_2O] = 2.004xx10^(-3) mi n^(-1)`

`k [55 mol L^(-1) ] = 2.004 xx10^(-3) mi n^(-1)`

`k = 3.64xx10^(-5) mol^(-1) L mi n^(-1)`