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, \(0.0048 \mathrm{~mol}\) \(\mathrm{PH}_{3}\) is consumed in a 2.0-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 P_4 and H_2 in this experiment are 0.0006 mol L^-1 s^-1 and 0.0036 mol L^-1 s^-1, respectively.

Step-by-step solution

Write down the balanced chemical equation

The balanced equation for the given reaction is: \( 4 \mathrm{PH}_{3}(g) \longrightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g) \)

Calculate the rate of disappearance of PH_3

The rate of disappearance of PH_3 is given as 0.0048 mol consumed per second in a 2.0 L container. Since rate is typically expressed in terms of concentration change (mol L^-1) over time, we can calculate the rate by dividing the amount consumed by the volume of the container: Rate of disappearance of PH_3 = \(\frac{0.0048 \ \text{mol}}{2.0 \ \text{L}\cdot \text{s}}\) = 0.0024 mol L^-1 s^-1

Relate the rate of disappearance of the reactant to the rate of appearance of the products

Using the stoichiometric coefficients in the balanced chemical equation, we can relate the rate of disappearance of PH_3 to the rate of appearance of P_4 and H_2. For every 4 moles of PH_3 consumed, 1 mole of P_4 and 6 moles of H_2 are produced. So, we can write: Rate of appearance of P_4 = (1/4) × Rate of disappearance of PH_3 Rate of appearance of H_2 = (6/4) × Rate of disappearance of PH_3

Calculate the rates of appearance of P_4 and H_2

Now, we can plug in the rate of disappearance of PH_3 and calculate the rates of appearance of P_4 and H_2: Rate of appearance of P_4 = (1/4) × 0.0024 mol L^-1 s^-1 = 0.0006 mol L^-1 s^-1 Rate of appearance of H_2 = (6/4) × 0.0024 mol L^-1 s^-1 = 0.0036 mol L^-1 s^-1 Thus, the rates of production of P_4 and H_2 in this experiment are 0.0006 mol L^-1 s^-1 and 0.0036 mol L^-1 s^-1, respectively.

Key concepts

Stoichiometry

When we talk about stoichiometry, we're diving into the heart of chemistry — the precise relationship between the amounts of reactants used and products formed in a chemical reaction. It's like a recipe where the 'ingredients' are elements and compounds and the 'end product' is an entirely new substance. The balanced chemical equation provides a ratio of how many moles of each substance are involved.

For the reaction in our exercise, \( 4 \text{PH}_3(g) \rightarrow \text{P}_4(g) + 6 \text{H}_2(g) \), stoichiometry offers the conversion factors to relate the consumption of \(\text{PH}_3\) to the production of \(\text{P}_4\) and \(\text{H}_2\). Specifically, the balanced equation tells us that for every four molecules of \(\text{PH}_3\) that react, one molecule of \(\text{P}_4\) is produced and six molecules of \(\text{H}_2\). By harnessing this information, it becomes possible to predict how much product will form from a certain amount of reactant, which is fundamental for everything from laboratory experiments to industrial chemical production.

Rate of Reaction

As fascinating as stoichiometry is, the story doesn't end there. Chemical reactions do not occur instantaneously, and this is where the 'rate of reaction' enters the stage. It is a measure of how quickly reactants turn into products. It's quantified by the change in concentration of a reactant or product over time, typically represented as \( \text{mol L}^{-1} \text{s}^{-1} \). Understanding this rate is crucial because it helps chemists control how fast a reaction proceeds to optimize processes like the manufacturing of chemicals.

In the example provided, by knowing the rate of disappearance of \(\text{PH}_3\), we are able to calculate the rates of appearance for \(\text{P}_4\) and \(\text{H}_2\) not just as abstract numbers, but as indicators of how fast the entire reaction is occurring. This information is invaluable for predicting the outcome of a reaction over time and is widely used in both academic and industrial settings.

Chemical Kinetics

Now let us bring in the final piece of the puzzle: chemical kinetics. This branch of chemistry delves into the factors that affect the rate of reaction and how the reaction rate can be manipulated. Kinetics gets to the 'why' and 'how' behind the pace at which a reaction advances. It includes the study of energy required to start a reaction (activation energy), the role of catalysts in tweaking reaction rates, temperature effects, concentration, and the actual mechanism by which reactants transform into products.

Through kinetics, concepts such as reaction rate can be fully appreciated because you're not just observing how fast a product forms but also understanding the conditions that are either accelerating or decelerating the reaction. In an educational setting, mastering chemical kinetics empowers students to not only follow set procedures in experiments but also to innovate by predicting and controlling reaction behaviors.