A second order reaction in which in both the reactants have same conentratio, is `20%` completed in 500 seconds. How much time it will take for `60%`
A second order reaction in which in both the reactants have same conentratio, is `20%` completed in 500 seconds. How much time it will take for `60%` completion?
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The second rorder equation when both the recants have same concentration is
`k=(1)/(t)*(x)/(a(a-x))`
if a=100, x=20 t=500 seconds.
So, `k=(1)/(500)xx(20)/(100(100-20))`
When `a=100, x=60 ,t=?`
`t=(1)/(lambda)*(60)/(100xx40)`
Substituting the value of K,
`t=(500xx100xx80)/(20)xx(60)/(100xx40)`
or t=3000 seconds
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