1.

Derive an expression for half life period of a first order reaction.

Answer»

Solution :We know that
`K=(2.303)/(t)log_(10)""([R]_(0))/([R])"…(1)"`
where, K = Rate constant t = TIME
`[R]_(0)=` Initial concentration
`[R]=` Concentration at any time
`"when"t=t_(1//2)" then"[R]=([R]_(0))/(2)`
Substitute these values in equation (1) we get
`K=(2.303)/(t_(1//2))LOG""(cancel([R]_(0)))/(cancel([R_(0)]_(0))/(2))`
`K=(2.303)/(t_(1//2))log2"[log 2 = 0.3010]"`
`K=(2.303)/(t_(1//2))xx0.3010`
`K=(0.693)/(t_(1//2))""rArr t_(1//2)=(0.693)/(K)`
For the FIRST order reaction, half life periodis independent of initial concentration of the reactants
i.e., `t_(1//2)alpha(0.693)/(K)`


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