Respuesta :
The full question can be seen below:
[tex]2H_{2}O_2_{(aq)} --> 2H_{2}O_{(l)}+O_{2}_{g}[/tex]
The decomposition of [tex]H_{2}O_{aq}[/tex] is represented by the equation above.
A student monitored the decomposition of a 1.0 L sample of [tex]H_{2}O_2_(_{aq}_)[/tex] at a constant temperature of 300K and recorded the concentration of [tex]H_{2}O_2[/tex] as function of time. The results are given in the table below:
                  Time (s)    [tex]H_{2}O_2[/tex]
                  0         2.7
                  200       2.1
                  400       1.7
                  600       1.4
The [tex]O_2_(_{g}_)[/tex] produced from the decomposition of the 1.0 L sample of [tex]H_{2}O_2_(_{aq}_)[/tex] is collected in a previously evacuated 10.0 L flask at 300 K. What is the approximate pressure in the flask after 400 s?
(For estimation purpose, assume that 1.0 mole of gas in 1.0 L exerts a pressure of 24 atm at 300 K).
Answer:
1.2 atm
Explanation:
Considering all assumptions as stated above;
            [tex]2H_{2}O_2_{(aq)} --> 2H_{2}O_{(l)}+O_{2}_{g}[/tex]
Initial        2.7 mole        ---       ---
Change      -1.0             ---       [tex]+\frac{1.0}{2}[/tex]
Equilibrium   1.7 mole         ---       0.5 mole
To determine the concentration of Oâ‚‚; we need to convert the moles to concentration for Oâ‚‚ = [tex]\frac{0.5}{volume in the flask}[/tex]
                 = [tex]\frac{0.5 mol}{10.0 L}[/tex]
                 = 0.05 [tex]\frac{mol}{L}[/tex]
Thus, based on the assumption that "1.0 mole of gas in 1.0 L exerts a pressure of 24 atm"
∴ 0.05[tex]\frac{mol}{L}[/tex] will give rise to = 0.05 [tex]\frac{mol}{L}[/tex] × 24
                      = 1.2 atm
The approximate pressure in the flask after 400 s is 1.2 atm
Calculation and Parameters:
Based on the given table
Initial        2.7 mole        ---       ---
Change      -1.0             ---       1/2
Equilibrium   1.7 mole         ---       0.5 mole
Hence, to find the concentration of Oâ‚‚; we need to convert the moles to concentration for Oâ‚‚ and this is done by 0.5/10L
=> 0.05 mol/L
Finally, we are acting on the assumption that "1.0 mole of gas in 1.0 L exerts a pressure of 24 atm" is 0.05 mol/L will then be 0.05 mol/L x 24
=> 1.2 atm
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