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Prove that the time required for the completion (3^(th))/(4) of the reaction of a fitst order is twice the time required for the completion of a half of the reaction. |
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Answer» SOLUTION :`t_((3)/(4))=(2.303)/(K)"log"([R]_(0))/((1)/(4)[R]_(6))` `t_((3)/(4))=(2.303)/(K)log 4` `= (2.303xx0.6021)/(K)=(1.386)/(K)` `= 2xx(0.693)/(K)` `= 2 t_((1)/(2))` |
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