Question

The decomposition of N_2O_5(g) following 1st order kinetics. N_2O_5(g) to N_2O_4(g) + ½ O_2(g) If 2.56 mg of N_2O_5 is initially present in a container and 2.50 mg remains after 4.26 min, what is the rate constant in 1/s?

Answer 1

Answer: The rate constant is
`0.334s^(-1)`

Explanation ;

Expression for rate law for first order kinetics is given by:

`t=(2.303)/(k)\log(a)/(a-x)`

where,

k = rate constant = ?

t = age of sample = 4.26 min

a = initial amount of the reactant = 2.56 mg

a - x = amount left after decay process = 2.50 mg

Now put all the given values in above equation to calculate the rate constant ,we get

`4.26=(2.303)/(k)\log(2.56)/(2.50)`

`k=(2.303)/(4.26)\log(2.56)/(2.50)`

`k=5.57* 10^(-3)min^(-1)=5.57* 10^(-3)* 60s^(-1)=0.334s^(-1)`

Thus rate constant is [tex]0.334s^{-1}