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A first order reaction is 25% complete in 40 minutes. Calculate the value of rate constant. In what time will the reaction be 80% completed?

Answer»

Solution :Apply the reaction
`k=2.303/t log ([R]_0)/([R])`
25% complete reactions means, `[R]=0.75[R]_0 and t=40 MINUTES`
Substituting the values in the equation above, we have
`k=2.303/(40 minutes)log""([R])_0/(0.7[R]_0)=2.303/(40 minutes)[log4-log 3]`
or `k=2.303/(40 minutes) (0.6021-0.4771)`
`or k=2.303/(40 minutes) TIMES 0.125=0.007197`
WRITING the first order equation again and subsituting new values, we have
`k=2.303/tlog""([R]_0)/([R])`
80% complete reaction means `[R]=0.2[R]_0`
Substituting the values in the equation above, we have
`0.007197=2.303/tlog""[R]_0/(0.2[R]_0)`
or `t=2.303/0.007197 timeslog 5=2.303/0.007197 times 0.699`
or `t=223.68 minutes`


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