The O2 produced from the decomposition of the 1.0 L sample of H2O2 is collected in a previously evacuated 10.0 L flask at 300. K What is the approximate pressure in the flask after 400 s

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

Read more about pressure here:

https://brainly.com/question/25736513

ACCESS MORE