1.

The rate cnstant k, for thereaction N_(2)O_(5)(g) rarr 2NO_(2)(g) + (1)/(2)O_(2)(g) is 23 xx 10^(-2) s^(-1). Which equation given below decribes the change of [N_(2)O_(5)] with time? [N_(2)O_(5)]_(0) and [N_(2)O_(5)]_(t) correspond to concentration of N_(2)O_(5) initially and at time, t

Answer»

`[N_(2)O_(5)] = [N_(2)O_(5)]_(0) + kt`
`[N_(2)O_(5)] = [N_(2)O_(5)]_(t)E^(kt)`
`log_(10) [N_(2)O_(5)]_(t) =log_(10) [N_(2)O_(5)]_(0) - kt`
`"ln"([N_(2)O_(5)]_(0))/([N_(2)O_(5)]t) = kt`

SOLUTION :Rate constant `= 2.3 xx 10^(-2) SEC^(-1)`
It means it is a first order reaction (because unit of rate constant is `sec^(-1)`)
For first order reaction`K = (1)/(t)ln(a)/(a-x)`
`kt = ln(a)/(a-x) = ln[N_(2)O_(5)]_(0)/([N_(2)O_(5)]_(t)`


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