Chemistry RATE LAW, RATE CONSTANT AND ITS UNIT

Rate law :

Experiment has shown that the rate of reaction mentioned above increases same number of times as the number of times the concentration of `N_2O_5` is increased. That is, rate is doubled by doubling the concentration of `N_2O_5`. This may be mathematically expressed as

Rate `mu[N_2O_5]` or Rate `=k[N_2O_5]`

Experiment has shown that the rate of reaction mentioned above increases same number of times as the number of times the concentration of `N_2O_5` is increased. That is, rate is doubled by doubling the concentration of `N_2O_5`. This may be mathematically expressed as

`-(d[N_2O_5])/(dt)=k [N_2O_5]`

`+(d[NO_2])/(dt)=k [N_2O_5]`

`-(d[O_2])/(dt)=k [N_2O_5]`

when `k` ` k` and `k` are the rate constants of the reaction. These three rate constants are inter-related.

`=>(k')/2=(k'')/4=k'''` as rate `= 1/2 {-(d[N_2O_5])/(dt)}=1/4{+(d[NO_2])/(dt)}=+(d[O_2])/(dt)`

Thus, for a reaction represented by the general equation

`aA+bB->cC+dD`

`1/a{-(dC_A)/(dt)}=1/b{-(dC_B)/(dt)}=1/c{-(dC_C)/(dt)}=1/d{-(dC_D)/(dt)}`

We have, `k^(I)/a=k^(II)/b=k^(III)/c=k^(IV)/d`

Where `k^(I), k^(II), k^(III)` and `k^(IV)` are the rate constants of the reaction when its rate is expressed in terms `A, B, C` and `D`, respectively.

Measurement of Rate of Reaction by Graphs :

(A) `text(Measurement of average rate of reaction)` :

To calculate the average rate of reaction between any two instants of time say `t_1` and `t_2`, the corresponding concentration `x_1` and `x_2` are noted from the graph. Then

Average rate of reaction `= (x_2-x_1)/(t_2-t_1)`

For example, from the fig.1, between the time interval ` 5` to `15` minutes,

Average rate `= (0.03-0.02)/(15-5)=(0.018)/10=0.0018` `mol L^(-1) min^(-1)`

(B) `text(Measurement of Instantaneous rate of reaction)` :

The rate of reaction at any time `t` is determined in the following way,

i) Concentration of any of the reactants or products whichever may be convenient is determined at various time intervals.

ii) Then concentration vs time curve is drawn.

iii) A tangent is drawn at the point `p` of the curve which corresponds to the time `t` at which rate is to be determined.

iv) The slope of the tangent gives the rate of reaction at the required time as shown in fig.2.

(`C_R` and `C_P` denote concentration of reactant and product respectively)

Unit of Rate of Reaction :

Unit of rate `=(text(Unit of Concentration))/(text(Unit of Time)) =text(Concentration time)^(- 1)`

i.e., `mol e//^(-1)s^(-1)`

For gaseous state it is also expressed in atm `s^(-1)` i.e. change in pressure per unit time.

Unit of Rate Constant :

The differential rate expression for `n^th` order reaction is as follows :

`(dx)/(dt)=k(a-x)^n` or `k = (dx)/((a-x)^n dt)=(text(concentration))/(text(concentration)^n time)=text(conc.)^(1-n) * text(time)^(-1)`

If concentration be expressed in mole`L^( - 1)` and time in minute, then `k =` `(mol e L ^(- 1))^(1-n) text(min)^(- 1)`

For zero order reaction: `n = 0` and hence `k = mol e L^( - 1) text(min)^(-1)`

For `1 st` order reaction: `n = 1` and hence `k =(mol e L^( - 1))^0 text(min)^(- 1) = text(min)^(- 1)`

For `2nd` order reaction: `n = 2` and hence `k =(mol e L^( - 1))^1 text(min)^(-1) = mol e^1 L text(min)^(- 1)`

The rate constant of a first order reaction has only time unit. It has no concentration unit. This means the numerical value of `k` for a first order reaction is independent of the unit in which concentration is expressed. If concentration unit is changed the numerical value of `k` for a first order reaction will not change. However, it would change with change in time unit. Say, `k` is `6.0 xx 10^(- 3)` `min^(-1)` then it may also be written as `1xx10^(-4) s^(-1)` i.e. numerical value of `k` will decrease `60` times if time unit is changed from however to minute or from minute to second.