Answer: The given question is incomplete. The complete question is:
At a certain temperature the rate of this reaction is second order in [tex]NH_4OH[/tex] with a rate constant of [tex]34.1M^{-1}s^{-1}[/tex] . [tex]NH_4OH(aq)\rightarrow NH_3(aq)+H_2O(aq)[/tex]
Suppose a vessel contains [tex]NH_4OH[/tex] at a concentration of 0.100 M Calculate how long it takes for the concentration of [tex]NH_4OH[/tex] to decrease to 0.0240 M. You may assume no other reaction is important. Round your answer to 2 significant digits.
Answer: It takes 0.93 seconds for the concentration [tex]NH_4OH[/tex] to decrease to 0.0240 M.
Explanation:
Integrated rate law for second order kinetics is given by:
[tex]\frac{1}{a}=kt+\frac{1}{a_0}[/tex]
[tex]a_0[/tex] = initial concentration = 0.100 M
a= concentration left after time t = 0.0240 M
k = rate constant = [tex]34.1M^{-1}s^{-1}[/tex]
t = time taken for decomposition = ?
[tex]\frac{1}{0.0240}=34.1\times t+\frac{1}{0.100}[/tex]
[tex]t=0.93s[/tex]
Thus it takes 0.93 seconds for the concentration [tex]NH_4OH[/tex] to decrease to 0.0240 M.
