Question

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

Answer 1

The full question can be seen below:

$2H_{2}O_2_{(aq)} --> 2H_{2}O_{(l)}+O_{2}_{g}$

The decomposition of $H_{2}O_{aq}$ is represented by the equation above.

A student monitored the decomposition of a 1.0 L sample of $H_{2}O_2_(_{aq}_)$ at a constant temperature of 300K and recorded the concentration of $H_{2}O_2$ as function of time. The results are given in the table below:

                                    Time (s)      $H_{2}O_2$

                                    0                 2.7

                                    200            2.1

                                    400            1.7

                                    600            1.4

The $O_2_(_{g}_)$ produced from the decomposition of the 1.0 L sample of $H_{2}O_2_(_{aq}_)$ 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;

                       $2H_{2}O_2_{(aq)} --> 2H_{2}O_{(l)}+O_{2}_{g}$

Initial               2.7 mole               ---             ---

Change          -1.0                        ---             $+\frac{1.0}{2}$

Equilibrium     1.7 mole                ---             0.5 mole

To determine the concentration of O₂; we need to convert the moles to concentration for O₂ = $\frac{0.5}{volume in the flask}$

                                  = $\frac{0.5 mol}{10.0 L}$

                                  = 0.05 $\frac{mol}{L}$

Thus, based on the assumption that "1.0 mole of gas in 1.0 L exerts a pressure of 24 atm"

∴ 0.05$\frac{mol}{L}$ will give rise to = 0.05 $\frac{mol}{L}$ × 24

                                           = 1.2 atm