Consider the reaction $$ 4 \mathrm{PH}_{3}(g) \longrightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g) $$ If, in a certain experiment, over a specific time period, \(0.0048\) mole of \(\mathrm{PH}_{3}\) is consumed in a \(2.0-\mathrm{L}\) container each second of reaction, what are the rates of production of \(\mathrm{P}_{4}\) and \(\mathrm{H}_{2}\) in this experiment?

The balanced chemical equation is given by: \[ 4 \mathrm{PH}_{3}(g) \longrightarrow \mathrm{P}_{4}(g) + 6 \mathrm{H}_{2}(g) \]

The question states that \(0.0048\) moles of \(\mathrm{PH}_{3}\) are consumed per second in a \(2.0-L\) container.

Observe the stoichiometric coefficients for the balanced equation: - For \(\mathrm{PH}_3\), it is \(4\) - For \(\mathrm{P}_4\), it is \(1\) - For \(\mathrm{H}_2\), it is \(6\)

To find the rate of production of \(P_4\), we will look at the relationship between the molar amounts of \(PH_3\) consumed and \(P_4\) formed. According to the balanced reaction, for every \(4\) moles of \(PH_3\) that react, \(1\) mole of \(P_4\) is formed. Therefore, we can set up the following proportion: \[ \frac{1 \, \text{mole of P}_4}{4\, \text{moles of PH}_3}= \frac{x \, \text{moles of P}_4}{0.0048\, \text{moles of PH}_3 \,\text{per second}} \] Now, solve for x: \[ x = \frac{1 \, \text{mole of P}_4}{4\, \text{moles of PH}_3} \times 0.0048\, \text{moles of PH}_3 \,\text{per second} = 0.0012\, \text{moles of P}_4 \,\text{per second} \] The rate of production of \(P_4\) is \(0.0012\) moles per second.

To find the rate of production of \(H_2\), we will look at the relationship between the molar amounts of \(PH_3\) consumed and \(H_2\) formed. According to the balanced reaction, for every \(4\) moles of \(PH_3\) that react, \(6\) moles of \(H_2\) are formed. Therefore, we can set up the following proportion: \[ \frac{6 \, \text{moles of H}_2}{4\, \text{moles of PH}_3}= \frac{y \, \text{moles of H}_2}{0.0048\, \text{moles of PH}_3 \,\text{per second}} \] Now, solve for y: \[ y = \frac{6 \, \text{moles of H}_2}{4\, \text{moles of PH}_3} \times 0.0048\, \text{moles of PH}_3 \,\text{per second} = 0.0072\, \text{moles of H}_2 \,\text{per second} \] The rate of production of \(H_2\) is \(0.0072\) moles per second.

The rate of production of \(P_4\) in this experiment is \(0.0012\) moles per second, and the rate of production of \(H_2\) is \(0.0072\) moles per second.