Question

The gas arsine, \(\mathrm{AsH}_{3},\) 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

The equilibrium pressure of \(\mathrm{H}_{2}(g)\) is 96.0 Torr and the equilibrium constant \(K_{\mathrm{p}}\) for this reaction is approximately 0.189.

Step-by-step solution

1. Set up the ICE table.

We will start by setting up an ICE table (Initial, Change, and Equilibrium) to keep track of the changes in pressure for each substance in the reaction. $$ \begin{array}{l||ccc} & 2 \mathrm{AsH}_{3}(g) & \rightleftharpoons & 2 \mathrm{As}(s) & + & 3 \mathrm{H}_{2}(g) \\ \hline \hline I & 392.0 & & 0 & & 0 \\ C & -2x & & +2x & & +3x \\ E & 392.0-2x & & 2x & & 3x \\ \end{array} $$ where x refers to the change in pressure.

2. Calculate change in pressure.

We know that the total pressure at equilibrium is 488.0 torr. We can express the change in pressure (\(x\)) in terms of the equilibrium pressures of the reactant and product gases: $$ (392.0 - 2x) + 3x = 488.0 $$ Solve for \(x\): $$ x = 32.0\,\text{Torr} $$

3. Calculate equilibrium pressure of \(\mathrm{H}_{2}(g)\).

Now that we know the value of \(x\), we can find the equilibrium pressure of the hydrogen gas by using the equilibrium row of our ICE table: $$ P_{\mathrm{H}_{2}} = 3x = 3(32.0) = 96.0\,\text{Torr} $$ So, the equilibrium pressure of \(\mathrm{H}_{2}(g)\) is 96.0 Torr.

4. Calculate equilibrium constant \(K_{\mathrm{p}}\).

We can now calculate the equilibrium constant \(K_{\mathrm{p}}\) using the equilibrium pressures of \(\mathrm{AsH}_{3}\) and \(\mathrm{H}_{2}\): $$ K_{\mathrm{p}} = \dfrac{P_{\mathrm{H}_{2}^3}}{(P_{\mathrm{AsH}_{3}})^2} = \dfrac{(96.0)^3}{(392.0 - 2 \times 32.0)^2} $$ Calculate \(K_{\mathrm{p}}\): $$ K_{\mathrm{p}} \approx 0.189 $$ So, the equilibrium constant \(K_{\mathrm{p}}\) for this reaction is approximately 0.189. In conclusion, the equilibrium pressure of \(\mathrm{H}_{2}(g)\) is 96.0 Torr and the equilibrium constant \(K_{\mathrm{p}}\) for this reaction is approximately 0.189.

Key concepts

ICE Table

An ICE table is a handy tool used in chemistry to track the concentrations or pressures of reactants and products during a reaction. The acronym ICE stands for Initial, Change, and Equilibrium. This table helps you understand and calculate how the concentrations or pressures vary from the start of the reaction to the point where the reaction reaches equilibrium.

In the given reaction, arsine (\(\text{AsH}_3\)) decomposes into solid arsenic and hydrogen gas. We start by determining the initial pressures, which is the starting condition before any reaction takes place. Here, the pressure of \(\text{AsH}_3\) is initially 392.0 torr, while the pressures of solid arsenic and hydrogen gas are zero because no reaction has occurred yet.

The next column in the ICE table is the "Change" which is expressed in terms of the variable \(x\). This variable represents the shift from the initial pressures to the equilibrium pressures, as dictated by the stoichiometry of the reaction. The pressures of reactants and products change as the reaction proceeds. For example, in this case, the pressure change for \(\text{AsH}_3\) is \(-2x\), while the change for hydrogen gas is \(+3x\).

Finally, the "Equilibrium" column displays the pressures at equilibrium, which are the result of adding the initial pressures and the changes. With this example, the equilibrium pressure for \(\text{AsH}_3\) is \(392.0 - 2x\), and the equilibrium pressure for \(\text{H}_2\) is \(3x\).

The ICE table provides a structured way of organizing information and simplifies the calculations needed to find out the system's behavior at equilibrium.

Equilibrium Constant (Kp)

The equilibrium constant, denoted as \(K_p\), represents the ratio of the pressures of products to reactants, each raised to the power of their coefficients in the balanced chemical equation, at equilibrium.

Calculating \(K_p\) is crucial as it helps us predict the extent of a reaction. A large \(K_p\) value implies the reaction favors products at equilibrium, while a small \(K_p\) suggests the reaction favors reactants.

To compute \(K_p\), we use the equilibrium pressures from our ICE table. In this reaction:

• Hydrogen, \(\text{H}_2\), has its equilibrium pressure represented as \(3x\)
• Arsine, \(\text{AsH}_3\), with the remaining pressure of \(392.0 - 2x\)

The equilibrium expression for this reaction is:\[K_p = \frac{P_{\text{H}_2}^3}{(P_{\text{AsH}_3})^2}\]

Substituting the known pressures into this expression lets us solve for \(K_p\), which in this scenario proves to be approximately 0.189. This value suggests that while some decomposition occurs, a significant amount of the original \(\text{AsH}_3\) remains at equilibrium.

Gas Pressure

Gas pressure is a crucial parameter in understanding the behavior of gases in chemical reactions, especially those involving equilibrium. Pressure is the force that the gas exerts per unit area on the walls of its container. It is often measured in units like atmospheres (atm) or torr.

In reactions involving gases, like the decomposition of arsine (\(\text{AsH}_3\)), pressure changes are indicative of the progress and equilibrium state of the reaction. Initially, the pressure of \(\text{AsH}_3\) is given, and as the reaction progresses, the pressure in the flask increases due to the formation of hydrogen gas (\(\text{H}_2\)).

In this problem, the setup includes an initial pressure of 392.0 torr for \(\text{AsH}_3\). Once equilibrium is reached, the observed total pressure is 488.0 torr. By understanding these pressure changes, and using the ICE table methodology, one can solve for the pressure contributions of different gases at equilibrium.

Gas pressure is thus essential not only for constructing the ICE table but also for applying it to create the equilibrium expression, thereby assisting in the calculation of equilibrium constants and understanding the shift in balance between reactants and products.