Answer :
Answer: It takes 5.89 s for the concentration of [tex]NH_3[/tex] to decrease by 88.0%.
Explanation:
[tex]2NH_3(g)\rightarrow N_2(g)+3H_2(g)[/tex]
As the units of rate constant is [tex]M^{-1}s^{-1}[/tex] , the kinetics must be second order.
Integrated rate law for second order kinetics is given by:
[tex]\frac{1}{a}=kt+\frac{1}{a_0}[/tex]
[tex]a_0[/tex] = initiaal concentration = 0.590 M
a= concentration left after time t = [tex]0.590-\frac{88}{100}\times 0.590=0.0708M[/tex]
[tex]\frac{1}{0.0708}=2.11\times t+\frac{1}{0.590}[/tex]
[tex]t=5.89s[/tex]
Thus it takes 5.89 seconds for the concentration of [tex]NH_3[/tex] to decrease by 88.0%.