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

Let's first write down the initial and equilibrium concentrations of all the species involved in the reaction. The reaction is: \(2 AsH_3 (g) \rightleftharpoons 2 As(s) + 3 H_2 (g)\) At the beginning: Pressure of AsH₃ = 392.0 torr Pressure of As (solid) = not applicable Pressure of H₂ = 0 torr (since initially there is no hydrogen gas present) After 48 hours - equilibrium is reached: Pressure of AsH₃ = 392.0 torr - 2x (since 2 moles of AsH₃ react) Pressure of As (solid) = not applicable Pressure of H₂= 3x torr (since 3 moles of H₂ are formed) The total pressure at equilibrium equals the sum of the pressures of all gaseous species: \(P_{total} = (392.0 - 2x) + 3x\) We are given the equilibrium pressure = 488.0 torr, so we can write: \(488.0 = 392.0 - 2x + 3x\)

We will solve the linear equation for x to find the change in pressure. \(488.0 = 392.0 - 2x + 3x\) \(488.0 = 392.0 + x\) Solving for x, we get: \(x = 488.0 - 392.0\) \(x = 96.0\)

Now that we have the value of x, we can find the equilibrium pressure of hydrogen gas (H₂). Pressure of H₂ at equilibrium = 3x \(Pressure_{H₂} = 3 * 96.0\) \:Pressure_{H₂} = 3 * 96.0 = 288.0 \: torr\) So, the equilibrium pressure of hydrogen gas is 288.0 torr.

Before calculating Kp value, we should find partial pressures at equilibrium for reacting AsH₃ molecule. Pressure of AsH₃ at equilibrium = 392.0 torr - 2x \(Pressure_{AsH₃} = 392.0 - 2 * 96.0\) \(Pressure_{AsH₃} = 392.0 - 192.0\) \(Pressure_{AsH₃} = 200.0 \: torr\)

Now we can calculate the equilibrium constant Kp using the partial pressures of the gaseous species at equilibrium. The expression for Kp is given by: \(K_p = \dfrac{(P_{H₂})^3}{(P_{AsH₃})^2}\) We can substitute the partial pressures we found earlier: \(K_p = \dfrac{(288.0)^3}{(200.0)^2}\) Now calculate Kp: \(K_p = \dfrac{24,157,056.0}{40,000}\) \(K_p = 603.9\)

a. The equilibrium pressure of H₂= 288.0 torr b. The value of Kp for the reaction = 603.9