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

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:

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