Question

At a certain temperature the rate of this reaction is second order in NH4OH with a rate constant of 34.1 M^-1*s^-1:
NH4OH(aq) -----> NH3(aq) + H2O (aq)
Suppose a vessel contains NH4OH at a concentration of 0.100M. Calculate how long it takes for the concentration of NH4OH to decrease to 0.0240 M. You may assume no other reaction is important. Round your answer to 2 significant digits.

Answer 1

Answer:

It will take 0.93 seconds the concentration of $NH_4OH$ to decrease to 0.0240 M.

Explanation:

Integrated rate law for second order kinetics is given by:

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

$t_{\frac{1}{2}=\frac{1}{k\times a_0}$

k = rate constant

$t_{1/2}$ = half life of reaction

$a_0$ = initial concentration

a= concentration left after time t

We have :

$NH_4OH(aq)\rightarrow NH_3(aq)+H_2O(aq)$

Initial concentration of ammonium hydroxide = $[NH_4OH]=a_o=0.100 M$

Final concentration of ammonium hydroxide after time t= $[NH_4OH]=a=0.0240 M$

Rate constant of the reaction = k = $34.1 M^{-1} s^{-1}$

$\frac{1}{0.0240 M}=34.1M^{-1} s^{-1}\times t+\frac{1}{0.100 M}$

Solving for t:

t = 0.93 seconds

It will take 0.93 seconds the concentration of $NH_4OH$ to decrease to 0.0240 M.