Question

The reaction AB(aq)→A(g)+B(g) is second order in AB and has a rate constant of 0.0164 M −1 ⋅ s −1 at 25.0 ∘ C . A reaction vessel initially contains 250.0 mL of 0.104 M AB which is allowed to react to form the gaseous product. The product is collected over water at 25.0 ∘ C . Part A How much time is required to produce 142.0 mL of the products at a barometric pressure of 707.3 mmHg . (The vapor pressure of water at this temperature is 23.8 mmHg .)

Answer 1

The reaction is second order in AB, so: $v=k[AB]^2$. In the statement, we obtain that $[AB]=0.104~M$ and, at 25 ºC, $k=0.0164~M^{-1}\cdot s^{-1}$. Then:

$v=k[AB]^2\\\\ v=0.0164\cdot0.104^2\\\\ v=0.0164\cdot0.010816\\\\ v\approx0.000177=1.77\times10^{-4}~mol/s$

Now, we'll calculate the number of mols of the products in the gas. Using the Ideal Gas Law:

$\bullet~\text{Pressure:}~p=707.3-23.8=683.5~mmHg\\\\ \bullet~\text{Volume:}~V=142~mL=0.142~L\\\\ \bullet~\text{Number of moles:}~n=n_A+n_B\\\\ \bullet~\text{Ideal gas constant:}~R=62.3~L\cdot mmHg\cdot K^{-1}\cdot mol^{-1}\\\\ \bullet~\text{Temperature:}~T=25^oC=25+273~K=298~K$

$pV=nRT\\\\ 683.5\cdot0.142=n\cdot62.3\cdot298\\\\ n=\dfrac{683.5\cdot0.142}{62.3\cdot298}\\\\ n\approx0,0052~mol$

Since each AB molecule forms one of A and one of B, $n_A=n_B$. Hence: $2n_A\approx0,0052\Longrightarrow n_A=n_B\approx0.0026~mol$.

We'll consider that in the beginning there was not A or B. So, $\Delta n_A=\Delta n_B=0.0026-0=0.0026~mol$. Furthermore, since the ratio of AB to A and to B is 1:1, $|\Delta n_{AB}|=|\Delta n_A|=|\Delta n_B|$.

Calculating the time by the expression of velocity:

$v=\dfrac{|\Delta[AB]|}{\Delta t}=\dfrac{1}{\Delta t}\cdot\dfrac{|\Delta n_{AB}|}{V}=\dfrac{1}{\Delta t}\cdot\dfrac{|\Delta n_A||}{250~mL}\\\\ 1.77\cdot10^{-4}=\dfrac{1}{\Delta t}\cdot\dfrac{0.0026~mol}{0.25~L}\\\\ \Delta t=\dfrac{0.0026}{0.25\cdot1.77\cdot10^{-4}}\\\\ \boxed{\Delta t\approx58.76~s}$