Question

For a reaction of second order , `t_(75%)=xt_(50%)`. The value of X is:

Answer 1

For the second order reaction:
`t=1/k[1/[[A]]-1/[A]_(0)]`
For `t_(1//2)` or `t_(50%)`
`t_(50%) = 1/k[(1/[A]_(0))/2-1/[A]_(0)]=1/k[2/[A]_(0)-1/[A]_(0)]`
`=1/k xx 1/[A]_(0)`
For `t_(3//4)` or `t_(75%)`
`t_(75%) = 1/k[(1/[A]_(0))/4-1/[A]_(0)]=1/k[4/[A]_(0)-1/[A]_(0)]=1/k xx 3/[A]_(0)`
`t_(75%) = x t_(50%)`
`t_(75%)/(t_(50%))=(1/k xx 3/[A]_(0))/(1/k xx 1/[A]_(0))=3`
`therefore x=3`