Question

The gas arsine, AsH_, decomposes as follows:$$2 \mathrm{AsH}_{3}(g) \rightleftharpoons 2 \mathrm{As}(s)+3 \mathrm{H}_{2}(g)$$.In an experiment at a certain temperature, pure \(\mathrm{AsH}_{3}(g)\) was placed in an empty, rigid, sealed flask at a pressure of 392.0 torr.After 48 hours the pressure in the flask was observed to be constant at 488.0 torr. a. Calculate the equilibrium pressure of \(\mathrm{H}_{2}(g)\) b. Calculate \(K_{\mathrm{p}}\) for this reaction.

Short answer

a. The equilibrium pressure of H2 is 144 torr. b. Kp for this reaction is 0.0299.

Step-by-step solution

Write the balanced equation

First, we write the balanced equation for the given decomposition: \[2 \mathrm{AsH}_{3}(g) \rightleftharpoons 2 \mathrm{As}(s) + 3 \mathrm{H}_{2}(g)\]

Identify initial and final pressures

According to the problem, the initial pressure of AsH3 is 392.0 torr, and the equilibrium pressure of all species in the flask is 488.0 torr.

Find change in pressure of AsH3 during the reaction

From the initial pressure and the final pressure, we can find the change in pressure (ΔP) of AsH3 during the reaction. ΔP = Initial pressure - Equilibrium pressure of AsH3 ΔP = 392 - 488 ΔP = -96

Calculate the change in pressure of H2

According to the balanced equation, 3 moles of H2 are produced from 2 moles of AsH3. So, we can easily calculate the change in pressure of H2. ΔP(H2) = \(\frac{3}{2}\) × ΔP(AsH3) ΔP(H2) = \(\frac{3}{2}\) × (-96) ΔP(H2) = -144

Calculate the equilibrium pressure of H2

The initial pressure of H2 is 0, the change in pressure of H2 is -144 torr. Hence, the equilibrium pressure of H2 is: P(H2) = Initial pressure + ΔP(H2) P(H2) = 0 + (-144) P(H2) = 144 torr

Calculate the equilibrium constant (Kp)

To calculate Kp, we can use the following expression: Kp = \(\frac{[H2]^3}{[AsH3]^2}\) Here, [H2] and [AsH3] refer to the equilibrium pressures in atm, not torr, so we need to first convert the pressures from torr to atm: P(AsH3) = 488 torr × \(\frac{1}{760}\) atm/torr = 0.6421 atm P(H2) = 144 torr × \(\frac{1}{760}\) atm/torr = 0.1895 atm Now, substitute the values into the Kp expression: Kp = \(\frac{(0.1895)^3}{(0.6421)^2}\) Kp = 0.0299 #Output# a. The equilibrium pressure of H2 is 144 torr. b. Kp for this reaction is 0.0299.

Key concepts

Chemical Equilibrium

When we talk about chemical equilibrium, we refer to a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentration—or in the case of gases, the pressure—of the reactants and products remains constant. It's essential to understand that this doesn't mean the reactants and products are present in equal amounts. Rather, their proportions are stable over time, which is what we observe as equilibrium.

To visualize this, imagine a busy street where the number of people entering a store is the same as those leaving it. Inside, the number of customers remains steady over time, much like how the pressures of reactants and products stay constant at equilibrium in a sealed container. This concept is crucial for comprehending phenomena in physical chemistry, such as the behavior of gases during a reaction.

Pressure Change in Reactions

When dealing with pressure change in reactions, particularly gas reactions, understanding the role of pressure is crucial. In a closed system, when a reaction involving gases takes place, the pressure can change due to the production or consumption of gaseous products or reactants.

For instance, if more gas moles are produced, the pressure within the container typically increases. Conversely, if more gas moles are used up than produced, there's a decrease in pressure. This phenomenon aligns with Avogadro's law, which states that, at constant temperature and volume, the number of moles of a gas is directly proportional to its pressure. In our example, the pressure change helps infer the progress of the reaction, indicating how much gas has been consumed or produced as the reaction reaches equilibrium.

Equilibrium Constant (Kp)

The equilibrium constant (Kp) is a value that expresses the ratio of the pressures of the products to the pressures of the reactants at equilibrium, each raised to the power of their coefficient in the balanced chemical equation. The 'p' in Kp denotes that we are dealing with pressures, making it particularly relevant for gas-phase reactions.

The calculation of Kp is not as simple as plugging in values; one must first ensure the pressures are in the correct units, typically atmospheres (atm). Kp is dimensionless and provides insight into the extent of the reaction: a large Kp suggests a reaction favoring the products, while a small Kp signals a reaction that favors the reactants. In educational contexts, it's important to clarify that Kp is determined solely by temperature—changes in pressure do not affect Kp directly, though they may shift the equilibrium position.