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In case of zero order reactions

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`t_(1//2)=2t_(1//4)`
`t_(3//4)=3t_(1//2)`
`t_(OO)=1t_(1//2)`
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Solution :`(C_(0)-C_(t))=x=Kt,(C_(0)-(C_(0))/2)=K.t_(1/2)impliest_(1/2)=(C_(0))/(2K),(C_(0)-(C_(0))/4)=K.t_(3/4)impliest_(3/4)=(3C_(0))/(4K)=3/2xx(C_(0))/(2K)`
`(C_(0)-(3C_(0))/4)=K.t_(1/4)impliest_(1/4)=(C_(0))/(4K)=1/2xx(C_(0))/(2K),C_(0)-0=Kt_(alpha)impliest_(alpha)=(C_(0))/K`
So `t_(3/4)=3/2xxt_(1/2),t_(1/4)=1/2xxt_(1/2),t_(1/2)=((t_(alpha))/2)`


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