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

Short answer

The rates of production of P₄ and H₂ in this experiment are \(0.0012\ mol\cdot s^{-1}\) and \(0.0072\ mol\cdot s^{-1}\), respectively.

Step-by-step solution

Identify the balanced reaction equation

We are given the balanced reaction equation: $$ 4 \mathrm{PH}_{3}(g) \longrightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g) $$

Determine the consumption rate of PH₃

We are given the consumption rate of PH₃ as 0.0048 moles per second.

Determine the relationship between rates

Using the stoichiometry of the balanced reaction equation, we can write down the relationships of the rates of consumption and production: $$ \frac{\text{rate of consumption of PH}_{3}}{4} = \frac{\text{rate of production of P}_{4}}{1} = \frac{\text{rate of production of H}_{2}}{6} $$

Find the rate of production of P₄

Dividing the rate of consumption of PH₃ by 4 gives us the rate of production of P₄: $$ \text{rate of production of P}_{4} = \frac{\text{rate of consumption of PH}_{3}}{4} = \frac{0.0048\ mol\cdot s^{-1}}{4} = 0.0012\ mol\cdot s^{-1} $$

Find the rate of production of H₂

Multiplying the rate of production of P₄ by 6 gives us the rate of production of H₂: $$ \text{rate of production of H}_{2} = 6 \times \text{rate of production of P}_{4} = 6 \times 0.0012\ mol\cdot s^{-1} = 0.0072\ mol\cdot s^{-1} $$

Final Answer

The rates of production of P₄ and H₂ are 0.0012 mol/s and 0.0072 mol/s, respectively.

Key concepts

Stoichiometry

Stoichiometry is a cornerstone concept in chemistry that deals with the quantitative relationships between the amounts of reactants and products in a chemical reaction. It is like a recipe for a chemical reaction, dictating exactly how much of each ingredient (or reactant) is needed and how much product will be made. This is crucial for experiments as it ensures that reactions are not only safe and efficient but also predictable.

In a chemical reaction, stoichiometry uses balanced chemical equations to describe these relationships. The coefficients in the balanced equation tell us the molar ratio of the reactants and products. For example, in the reaction \(4 \mathrm{PH}_{3}(g) \rightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g)\):

• 4 moles of \(\mathrm{PH}_3\) produce 1 mole of \(\mathrm{P}_4\)
• 4 moles of \(\mathrm{PH}_3\) produce 6 moles of \(\mathrm{H}_2\)

Therefore, if we know how fast \(\mathrm{PH}_3\) is being consumed (in moles per second), we can calculate how fast \(\mathrm{P}_4\) and \(\mathrm{H}_2\) are being produced using these ratios.

Stoichiometry gives us the framework to calculate these rates, making it a vital tool in chemical engineering, research, and industry.

Chemical Reactions

A chemical reaction involves the transformation of one or more substances into new substances. This magical transformation is what changes \(\mathrm{PH}_3\) into \(\mathrm{P}_4\) and \(\mathrm{H}_2\) in our example reaction. During this process, bonds between atoms in the reactants are broken, and new bonds are formed to create the products.

This specific reaction is a decomposition reaction where a single compound breaks into two or more simpler products. Recognizing the type of chemical reaction helps in understanding how materials interact with each other and what energy changes are taking place.

The speed at which a reaction occurs is referred to as the reaction rate. It can be influenced by different factors including temperature, concentration of reactants, surface area, and the presence of a catalyst. In the given reaction, the rate is measured by the disappearance of \(\mathrm{PH}_3\) per second and the simultaneous emergence of \(\mathrm{P}_4\) and \(\mathrm{H}_2\). This interconnectedness emphasizes the beauty and complexity of chemical reactions in action.

Balanced Equations

Balanced equations are essential in chemistry to ensure that the same number of each type of atom appears on both sides of a chemical equation. This mirrors the Law of Conservation of Mass, which states that mass cannot be created or destroyed in a chemical reaction.

In our reaction example, the balanced equation \(4 \mathrm{PH}_{3}(g) \rightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g)\) ensures that all phosphorus and hydrogen atoms from the reactants are accounted for among the products.

When balancing equations, it's important to adjust the coefficients (the numbers in front of the molecules) rather than changing the chemical formulas themselves. This preserves the identity of the chemicals involved while maintaining atom balance.

• The 4 in front of \(\mathrm{PH}_3\) indicates we need four \(\mathrm{PH}_3\) molecules.
• The 1 in front of \(\mathrm{P}_4\) (though often omitted) indicates one is produced.
• The 6 in front of \(\mathrm{H}_2\) shows six \(\mathrm{H}_2\) molecules are produced.

This systematic approach not only helps predict how much of each substance is needed and produced but also allows chemists to precisely control chemical reactions in laboratory and industrial settings.