Question

At a certain temperature the rate of this reaction is second order in with a rate constant of Suppose a vessel contains at a concentration of . Calculate how long it takes for the concentration of to decrease to . You may assume no other reaction is important. Round your answer to significant digits.

Answer 1

Answer: The given question is incomplete. The complete question is:

At a certain temperature the rate of this reaction is second order in $NH_4OH$ with a rate constant of $34.1M^{-1}s^{-1}$ . $NH_4OH(aq)\rightarrow NH_3(aq)+H_2O(aq)$

Suppose a vessel contains $NH_4OH$ at a concentration of 0.100 M Calculate how long it takes for the concentration of $NH_4OH$  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  $NH_4OH$  to decrease to 0.0240 M.

Explanation:

Integrated rate law for second order kinetics is given by:

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

$a_0$ = initial concentration = 0.100 M

a= concentration left after time t = 0.0240 M

k = rate constant = $34.1M^{-1}s^{-1}$

t = time taken for decomposition = ?

$\frac{1}{0.0240}=34.1\times t+\frac{1}{0.100}$

$t=0.93s$

Thus it takes 0.93 seconds  for the concentration  $NH_4OH$  to decrease to 0.0240 M.