When 5.1 g of solid NH HS is introduced into a two litre evacuated flask at 27°C, 20% of the solid d — Chemical Equilibrium Chemistry Question
Question
When 5.1 g of solid NH HS is introduced into a two litre evacuated flask at 27°C, 20% of the solid decomposes into gaseous ammonia and hydrogen sulphide. The Kp for the reaction at 27°C is x × 10 . The value of x is _______. (Integer answer) [Given R = 0.082 L atm K mol ] 4 –2 –1 –1
💡 Solution & Explanation
# Solution: Calculating Kp for NH₄HS Decomposition **Step 1: Write the reaction and find moles of NH₄HS** NH₄HS(s) ⇌ NH₃(g) + H₂S(g) Molar mass of NH₄HS = 14 + 4 + 32 + 1 = 51 g/mol Moles of NH₄HS = 5.1/51 = 0.1 mol **Step 2: Determine moles of gases produced** 20% decomposition: 0.1 × 0.20 = 0.02 mol decomposed From stoichiometry: 0.02 mol NH₃ and 0.02 mol H₂S produced **Step 3: Calculate partial pressures** Using PV = nRT: P(NH₃) = (0.02 × 0.082 × 300)/2 = 0.246 atm P(H₂S) = (0.02 × 0.082 × 300)/2 = 0.246 atm **Step 4: Calculate Kp** Kp = P(NH₃) × P(H₂S) = 0.246 × 0.246 = 0.0605 atm² **Step 5: Express in the given form** Kp = 0.0605 = 6.05 × 10⁻² ≈ 6 × 10⁻² Therefore, x = 6, and the answer is **6.00**.