1.

A first order reaction is 40% complete in 80 minutes. Calculate the value of rate constant (k). In what time will the reaction be 90% completed ? [Given : log2=0.3010,log3=0.4771,log4=0.6021,log5=0.6771,log6=0.7782]

Answer»

Solution :Equation for first order reaction is
`K=(2.303)/(t)log([R]_(0))/([R])`
For 40% complete reaction, [R] = 0.6 `[R]_(0)`, t = 80 minutes SUBSTITUTING the values in the above equation, we have
`k=(2.303)/(80)log([R]_(0))/(0.6[R]_(0))`
or `""k=(2.303)/(80)[log5-log3]=(2.303)/(80)[0.6771-0.4771]`
or `""k=(0.4606)/(80)=0.005758`
Now, `""0.005758=(2.303)/(t)log([R]_(0))/(0.1[R]_(0))=(2.303)/(t)XX1`
or `""t=(2.303)/(0.005758)=400` minutes


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