The reaction : `2N_(2)O_(5)(g) to 2NO_(2)(g) + (O_(2))_(g)`
was studied and the following data were collected.

Determine: (i) the order, (ii) the rate law and (iii) rate constant for the reaction.

Explanation

Let the order of the reaction = n
`Rate propto [N_(2)O_(5)]^(n)`
`(34 xx 10^(-5)) propto (1.13 xx 0^(-2))^(n)`
`(25 xx 10^(-5)) propto (0.84 xx 10^(-2))^(n)`
Dividing eqn. (i) by eqn. (ii), we have
`(34 xx 10^(-5))/(25 xx 10^(-5)) = (1.13 xx 10^(-2))^(n)/(0.84 xx 10^(-2))^(n), (1.36)^(1) = (1.35)^(n)` or n=1
i) The order of the reaction = 1,
ii) Rate law, The rate of reaction depends upon one concentration team of `N_(2)O_(5)` because of the reaction is 1.
ii) Rate law. The rate of reaction depends upon one concentration term of `N_(2)O_(5)` because order of the reaction is 1.
Reaction Rate = `k[N_(2)O_(5)]`
iii) Rate constant, `k = "Reaction rate"/([N_(2)O_(5)])= (18 xx 10^(-5) mol L^(-1)min^(-1))/(0.62 xx 10^(-2)mol L^(-1))= 0.0.29 min^(-1)`