Question

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 \(\mathbf{P}_{4}\) and \(\mathbf{H}_{2}\) in this experiment?

Short answer

In this experiment, the rates of production of \(P_4\) and \(H_2\) are 0.0012 moles per second and 0.0072 moles per second, respectively.

Step-by-step solution

Write down the given information

We are given that the rate of consumption of PH3 is 0.0048 moles per second in a 2.0-L container.

Note the stoichiometric relationship from the balanced equation

Using the balanced chemical equation, we can determine the stoichiometric relationship between PH3, P4, and H2. The balanced chemical equation is: \(4 \, \text{PH}_3(g) \longrightarrow \text{P}_4(g) + 6 \, \text{H}_2(g)\) - For every 4 moles of PH3 consumed, 1 mole of P4 is produced and 6 moles of H2 are produced.

Calculate the rate of production of P4

Since the stoichiometric relationship states that 1 mole of P4 is produced for every 4 moles of PH3 consumed, we can divide the given rate of consumption of PH3 by 4 to find the rate of production of P4: Rate of production of P4 = (0.0048 moles of PH3 consumed per second) × (1 mole of P4/4 moles of PH3) Rate of production of P4 = 0.0012 moles per second

Calculate the rate of production of H2

Following the same procedure, using the stoichiometric relationship, 6 moles of H2 are produced for every 4 moles of PH3 consumed. Therefore, to find the rate of production of H2, we can multiply the given rate of consumption of PH3 by 6/4: Rate of production of H2 = (0.0048 moles of PH3 consumed per second) × (6 moles of H2/4 moles of PH3) Rate of production of H2 = 0.0072 moles per second

Conclusion

In this experiment, the rates of production of P4 and H2 are as follows: - P4: 0.0012 moles per second - H2: 0.0072 moles per second

Key concepts

Stoichiometry

Understanding stoichiometry is critical when working with chemical reactions. It refers to the calculation of reactants and products in chemical reactions. It is the mathematical relationship between the quantities of reactants and products. Stoichiometry is based on the conservation of mass where the total mass of the reactants equals the total mass of the products.

To apply stoichiometry to a problem, you first need a balanced chemical equation, which shows the proportions of reactants and products. For instance, in the given exercise, the balanced equation is \(4 \mathrm{PH}_{3}(g) \longrightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g)\). This tells us that four moles of phosphine gas (\(\mathrm{PH}_{3}\)) react to produce one mole of tetraphosphorus (\(\mathrm{P}_{4}\)) and six moles of hydrogen gas (\(\mathrm{H}_{2}\)).

When we calculate reaction rates using stoichiometry, we compare the rate at which the reactants decrease to the rate at which the products are formed. The rates at which reactants or products are formed or consumed can be expressed in terms of molarity, which is mole per volume (e.g., mol/L), or as in our exercise, moles per second when considering the time factor.

Balanced Chemical Equations

Balanced chemical equations are integral to chemistry because they ensure the Law of Conservation of Mass is upheld. In a balanced equation, the number of atoms for each element is the same on both the reactants and products side. This balancing act is crucial for making stoichiometric calculations.

For example, the balanced equation from our exercise, \(4 \mathrm{PH}_{3}(g) \longrightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g)\), shows that four phosphorus atoms are present in both reactants and products, and twelve hydrogen atoms are balanced on both sides. The coefficients in a balanced equation also tell us the stoichiometric ratios, which we use to calculate the amount of products formed from a given amount of reactants. This ratio was key to solving the exercise, as it guides the conversion between the consumption of \(\mathrm{PH}_{3}\) and the production rates of \(\mathrm{P}_{4}\) and \(\mathrm{H}_{2}\).

Rate of Production

The rate of production in a chemical reaction refers to the speed at which product molecules are formed. It can be considered the 'output speedometer' for a chemical reaction. The rate of production varies for different substances within a reaction depending on their stoichiometric coefficients in the balanced chemical equation.

In our exercise, we were asked to calculate the rate of production of \(\mathrm{P}_{4}\) and \(\mathrm{H}_{2}\) based on the known consumption rate of \(\mathrm{PH}_{3}\). Using stoichiometry, the rate of production for \(\mathrm{P}_{4}\) is \frac{1}{4}th of the rate of consumption of \(\mathrm{PH}_{3}\) because for every four moles of \(\mathrm{PH}_{3}\) that react, only one mole of \(\mathrm{P}_{4}\) is produced. Similarly, the rate of production for \(\mathrm{H}_{2}\) is \frac{6}{4} times that of \(\mathrm{PH}_{3}\)'s consumption rate, as six moles of \(\mathrm{H}_{2}\) are produced for every four moles of \(\mathrm{PH}_{3}\) that react.

Thus, the role of stoichiometry in determining the rate of production is pivotal. Calculating these rates properly requires a clear understanding of the balanced chemical equation and the stoichiometric relationships, which together guide the conversion from reactant consumption to product formation rates.