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 \(\operatorname{AsH}_{3}(g)\) 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 \(\mathrm{H}_{2}(\mathrm{~g})\). b. Calculate \(K_{\mathrm{p}}\) for this reaction.

Short answer

The equilibrium pressure of H2(g) is found to be 144 torr, and the equilibrium constant Kp for this reaction is 0.163.

Step-by-step solution

a. Calculate the equilibrium pressure of H2(g)

The initial pressure of AsH3 is 392.0 torr. After reaching equilibrium, the total pressure (including H2) is 488.0 torr. Let's denote the decrease in pressure of AsH3 as x. Since 2 moles of AsH3 form 3 moles of H2, the pressure of H2 formed will be 3x/2. From the total pressure, we can write: Initial Pressure_(AsH3) - x + (3x/2) = Final Pressure 392.0 - x + (3x/2) = 488.0 Now, solve for x.

b. Solve for x

Solving the equation 392.0 - x + (3x/2) = 488.0, we get: 3x/2 - x = 488 - 392 3x - 2x = 2(96) x = 96 torr So, the change in pressure of AsH3 is 96 torr. Now we will find the equilibrium pressure of H2.

c. Find the equilibrium pressure of H2

We know that 2 moles of AsH3 form 3 moles of H2. So, if the pressure of AsH3 decreased by 96 torr, the pressure of H2 will be: Pressure_(H2) = (3/2) * 96 = 144 torr The equilibrium pressure of H2 is 144 torr.

d. Calculate Kp for the reaction

Now that we have the equilibrium pressures of the reactants and products, we will calculate the equilibrium constant, Kp. Kp = [Pressure_(H2)^3] / [Pressure_(AsH3)^2] The equilibrium pressure of AsH3 is (Initial Pressure - x) = (392 - 96) = 296 torr. Now we can calculate Kp: Kp = (144^3) / (296^2)

e. Find the value of Kp

Calculating Kp for the reaction, we get: Kp = (144^3) / (296^2) = 0.163 The equilibrium constant, Kp, for this reaction is 0.163.

Key concepts

Equilibrium Pressure

Equilibrium pressure refers to the pressure of a gas when a chemical reaction reaches a state where the rates of the forward and reverse reactions are equal.
This means that the amount of reactant turning into product is equal to the amount of product reverting back into the reactant, resulting in no net change in the concentrations of reactants and products over time.
In the case of the decomposition of arsine (\(\mathrm{AsH}_3\)), the equilibrium pressure is the pressure measured in the flask after the system reaches equilibrium. - Initially, only the pressure of pure \(\mathrm{AsH}_3\) is considered, which was \(392.0\) torr. - As the reaction progresses towards equilibrium, this pressure shifts to \(488.0\) torr, which includes the contribution from the formed gas \(\mathrm{H}_2\) as well.Thus, at equilibrium, the pressure of \(\mathrm{H}_2\) is calculated as \(144\) torr, indicating the amount of gas formed as a result of the decomposition reaction reaching a balance.

Equilibrium Constant (Kp)

The equilibrium constant, denoted as \(K_p\), is a central concept in understanding chemical equilibria in gaseous systems.
It is defined based on the partial pressures of the gases involved in the reaction at equilibrium.The relationship for calculating \(K_p\) in a reaction is given by the expression derived from the balanced chemical equation, where the pressures of products are divided by the pressures of reactants.
For the decomposition of \(\mathrm{AsH}_3\), this is expressed as:\[K_p = \frac{(\text{Pressure of } \mathrm{H}_2)^3}{(\text{Pressure of } \mathrm{AsH}_3)^2}\]
In our case:- \(\mathrm{H}_2\) reached an equilibrium pressure of \(144\) torr.- The remaining \(\mathrm{AsH}_3\) pressure is \(296\) torr (calculated as \(392 - 96\) torr).Substituting these values into the equation gives:\[K_p = \frac{144^3}{296^2} = 0.163\] This \(K_p\) value indicates the extent to which a reaction favors the formation of products at equilibrium under given conditions.

Decomposition Reaction

A decomposition reaction is a type of chemical reaction where a single compound breaks down into two or more simpler substances.
It usually requires energy input, like heat, light, or electricity, to proceed.In the reaction involving arsine (\(\mathrm{AsH}_3\)), the gas decomposes into solid arsenic (\(\mathrm{As}\)) and hydrogen gas (\(\mathrm{H}_2\)).- The initial compound, \(\mathrm{AsH}_3\), undergoes a chemical change, evidenced by the decrease in its pressure as it loses energy.- The products formed are more stable at the reaction conditions provided, leading to the observed equilibrium of gases in the system.This type of reaction is important in understanding how complex substances can break into simpler components, shifting the balance of reactions and affecting the equilibrium constants and pressures in chemical systems.