Solid ammonium hydrogen sulfide is introduced into a 2.00-L flask, and the flask is sealed. If this solid decomposes according to the equation below
NH₄HS(s) ⇌ NH₃(g) + H₂S(g) , K_p = 0.108 at 25^$\circ$C,
What is the minimum mass of ammonium hydrogen sulfide that must be present in the flask initially if equilibrium is to be established at 25^$\circ$C

Question

Quizplus · Accepted Answer

The equilibrium constant $K_p = 0.108$ for the decomposition of ammonium hydrogen sulfide ($NH_4HS$) into ammonia ($NH_3$) and hydrogen sulfide ($H_2S$) gases. At equilibrium, the partial pressures of $NH_3$ and $H_2S$ will be equal since they are produced in a 1:1 ratio. Let $x$ be the equilibrium partial pressure of each gas. Thus, $K_p = x^2$, and solving for $x$ gives $x = \sqrt{0.108}$. The total pressure at equilibrium is $2x$, and using the ideal gas law $PV = nRT$, where $n$ is the number of moles, $R$ is the gas constant (0.0821 L·atm/mol·K), and $T$ is the temperature in Kelvin (298K for 25°C), we can find the moles of gas produced. Since the molar mass of $NH_4HS$ is approximately 51 g/mol, calculating the mass of $NH_4HS$ that decomposes to reach equilibrium yields a value closest to the option given as 1.37 g.

Solid Ammonium Hydrogen Sulfide Is Introduced into a 2 ∘\circ∘ C, What Is the Minimum Mass of Ammonium Hydrogen Sulfide

Solid Ammonium Hydrogen Sulfide Is Introduced into a 2 $\circ$ C,
What Is the Minimum Mass of Ammonium Hydrogen Sulfide