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:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

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

$4.26=\frac{2.303}{k}\log\frac{2.56}{2.50}$

$k=\frac{2.303}{4.26}\log\frac{2.56}{2.50}$

$k=5.57\times 10^{-3}min^{-1}=5.57\times 10^{-3}\times 60s^{-1}=0.334s^{-1}$

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