1.

Rate constant of a first order reaction is 0.0693 min^(-1). Calculate the percentage of the reactant remaining at the end of 60 minutes.

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Solution :FORMULA : `K = (2.303)/(t) log""([R]_(0))/([R])`
` 0.0693 = (2.303)/(60) log""(100)/([R]) rArr 1.8 = log (100)/([R])`
ANTILOG `1.8 = (100)/([R])`
`63.1 = (100)/([R]) rArr [R] = 1.584%`


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