Question

Consider this reaction: →2NH3g+N2g3H2g At a certain temperature it obeys this rate law: rate =2.11·M−1s−1NH32 Suppose a vessel contains NH3 at a concentration of 0.590M. Calculate how long it takes for the concentration of NH3 to decrease by 88.0%. You may assume no other reaction is important.

Answer 1

Answer: It takes 5.89 s for the concentration of $NH_3$ to decrease by 88.0%.

Explanation:

$2NH_3(g)\rightarrow N_2(g)+3H_2(g)$

As the units of rate constant is $M^{-1}s^{-1}$ , the kinetics must be second order.

Integrated rate law for second order kinetics is given by:

$\frac{1}{a}=kt+\frac{1}{a_0}$

$a_0$ = initiaal concentration = 0.590 M

a= concentration left after time t = $0.590-\frac{88}{100}\times 0.590=0.0708M$

$\frac{1}{0.0708}=2.11\times t+\frac{1}{0.590}$

$t=5.89s$

Thus it takes 5.89 seconds for the concentration of $NH_3$ to decrease by 88.0%.