1.

The time for half life change in a first order decomposition of a substance A is 60 seconds. Calculate the rate constant. How much of A will be left after 180 seconds ?

Answer»

Solution :`t^((1)/(2))= 60` min
Order of reaction = first order, So, for first order reaction,
We have
`t^((1)/(2))=(0.693)/(K)`
`K=(0.693)/(t_((1)/(2)))=(0.693)/(60)=0.0115 S^(-1)`
`K=0.0115 S^(-1)`.
For TIME 3 mins (180 Sec)
`K=(2.303)/(t)"log"([a_(0)])/([a])`
`therefore 0.0115 = (2.303)/(t)"log"([a_(0)])/([a])`
`log = ([a_(0)])/([a])0.014`
`([a_(0)])/([a])` = ANTILOG (0.014)
`([a_(0)])/([a])=1.0327`
% of the reactant left after 3 min
`([a])/([a_(0)])xx100=(1)/(4)=25%`.


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