$$\ce{2H2O2(aq) -> 2H2O(l) + O2(g) } \hskip{3em}$$

The decomposition of $\ce{H2O2(aq)}$ is represented by the equation above. A student monitored the decomposition of a $\pu{1.0 L}$ sample of $\ce{H2O2(aq)}$ at a constant temperature of $\pu{300}$. $\pu{K}$ and recorded the concentration of $\ce{H2O2}$ as a function of time. The results are given in the table below.

$$\begin{array}{|c|c|} \hline \text{Time (s)} & \ce{[H2O2]} \\ \hline \hline 0 & 2.7 \\ \hline 200. & 2.1 \\ \hline 400. & 1.7 \\ \hline 600. & 1.4 \\ \hline \end{array}$$

The $\ce{O2(g)}$ produced from the decomposition of the $\pu{1.0 L}$ sample of $\ce{H2O2(aq)}$ is collected in a previously evacuated $\pu{10.0 L}$ flask at $\pu{300}$. $\pu{K}$. What is the approximate pressure in the flask after $\pu{400}$. $\pu{s}$ ? (For estimation purposes, assume that $\pu{1.0 }$mole of gas in $\pu{1.0 L}$ exerts a pressure of $\pu{24 atm}$ at $300$. $\pu{K}$.)

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We can use the ideal gas law equation. Start by finding the number of moles of oxygen through the table and reaction.

$$\ce{[H2O2]} = 2.7-1.7$$ $$= \pu{1.0 M } \ce{[H2O2]}$$ $$\frac{\pu{1 mol } \ce{H2O2}}{\pu{1 L} } \cdot \pu{1.0 L} \cdot \frac{\pu{1 mol } \ce{O2}}{\pu{2 mol }\ce{H2O2}}$$ $$= \pu{0.5 mol }\ce{O2}$$

Plugging in the values:

$$PV=nRT$$ $$P(10.0) = (0.5)(0.0821)(300)$$ $$P \approx \boxed{1.2}$$

We can use the note about estimation to set a proportion. Since temperature is held constant:

$$PV=nRT$$ $$\frac{PV}{n}=RT \hskip{1em} \frac{PV}{n}=k$$ $$\frac{P_1V_1}{n_1} = \frac{P_2V_2}{n_2}$$ $$\frac{(P_1)(10.0)}{0.5} =\frac{(24)(1.0)}{1.0}$$ $$P_1 = \boxed{1.2}$$