Saved Bookmarks
| 1. |
Derive the relation between the half-life period andrate constant of first order reaction. |
|
Answer» Solution : Consider the following reaction, `{:(,A to B),(,a""0),(,(a-X)""x):}` CONCENTRATION at time t = 0 Concentration at time t = t If `[A]_(0)` and `[A]_(t)` arethe CONCENTRATIONS of A at start and after time t,then `[A]_(0)= a` and`[A]_(t) = a -x` The velocity constant or the specific rate constant k for the first order reaction can be represented as, `k=(2.303)/(t) log_(10). ([A]_(0))/([A]_(t))` `thereforek = (2.303)/(t) log_(10).((a)/(a-x))` where, a is the initial concentration of the reactant A, x is the concentration of the product B after time t, so that (a - x) is theconcentration of the reactant A after time t. If `t_(1//2)`is the half-lifeon a reaction , TENAT `t = t_(1//2), x = a//2` , hencea - x = a - a/2 = a/2 Now, `k = (2.303)/(t) log_(10).((a)/(a-x))` `THEREFORE t = (2.303)/(k)log_(10).(a)/((a-x))` Hence, `t_(1//2) = (2.303)/(k) log_(10).(a)/(a//2)` ` = (2.303)/(k) log_(10)2` `= (2.303 xx 0.3010)/(k)` ` because t_(1//2) = (0.693)/(k)` |
|