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Show that for first order reaction t(87.5%) = 3t_(50%) |
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Answer» Solution :`k =2.303/t_(87.5%)"LOG" (100)/(12.5)``k=2.303/t_(87.5%)xx0.9031` EXPRESSION of K when 50% of REACTION is reacted `k=2.303/t_(50%) "log" (100)/(50)``k= 2.303/t_(50%)xx0.3010`k= EQUATION `(1)//`Equation(2) `t_(87.5%)=t_(50%)` |
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