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A first order reaction takes 10 minutes for 25% decomposition. Calculate t_(1//2) for the reaction. [Given : log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021]

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Solution :The equation for first ORDER reaction is
`t=(2.303)/(k)"log"([R]_(0))/([R])""…(1)`
For 25% decomposition, `[R]=(3[R]_(0))/(4)`
SUBSTITUTING the VALUES in equation (1), we have
`10=(2.303)/(k)"log"([R]_(0)xx4)/(3[R]_(0))`
or `10=(2.303)/(k) "log"(4)/(3)""...(2)`
For 50% decomposition, `[R]=([R]_(0))/(2)`
Substituting the values in equation (1), we have
`t_(1//2)=(2.303)/(k)"log"([R]_(0)xx2)/([R]_(0))`
or `t_(1//2)=(2.303)/(k)log2 ""...(3)`
Dividing (3) by (2), we have
`(t_(1//2))/(10)=(log2)/("log"(4)/(3)) or t_(1//2)=10xx(log2)/((log4-log3)) or t_(1//2) = 10xx (0.3010)/((0.6021-0.4771))`
or `t_(1//2)=(10xx0.3010)/(0.1250)=24` minutes


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