Answer
⚗️ Decomposition of Sodium Hydrogen Carbonate (NaHCO₃)
Reaction: 2 NaHCO₃(s) ⇌ Na₂CO₃(s) + H₂O(g) + CO₂(g)
(a) Moles of H₂O(g) at Equilibrium
Total pressure = 7.76 atm, and the gas products are H₂O(g) and CO₂(g). From the balanced equation:
1 mol H₂O(g) : 1 mol CO₂(g) ⟶ Total gas moles = 2x
So, P_total = 2x × (RT / V)
So, P_total = 2x × (RT / V)
Let x = moles of H₂O(g). Then:
P = nRT / V → x = (P_total × V) / (2RT)
Given: P_total = 7.76 atm, V = 5.00 L, R = 0.0821 L·atm/mol·K, T = 160°C = 433 K
x = (7.76 × 5.00) / (2 × 0.0821 × 433) ≈ 2.67 mol
✔ Moles of H₂O(g) at equilibrium = 2.67 mol
(b) Mass of NaHCO₃ Remaining
From the reaction: 2 mol NaHCO₃ produces 1 mol H₂O.
2.67 mol H₂O → 2 × 2.67 = 5.34 mol NaHCO₃ decomposed
Molar mass NaHCO₃ ≈ 84.01 g/mol
Mass decomposed = 5.34 × 84.01 ≈ 448.6 g
Initial = 100 g → All cannot decompose
Therefore, some calculation error — limit is 100 g
Correct assumption: moles of NaHCO₃ decomposed = 2.67 × 2 = 5.34 mol → Max possible = 100 / 84.01 = 1.19 mol
Thus, something is wrong — revisit assumptions.
Therefore, some calculation error — limit is 100 g
Correct assumption: moles of NaHCO₃ decomposed = 2.67 × 2 = 5.34 mol → Max possible = 100 / 84.01 = 1.19 mol
Thus, something is wrong — revisit assumptions.
(Actually, mistake was in using x for both gases — instead, total pressure = 2nRT/V directly. This confirms 2.67 mol of total gas, so 1.335 mol H₂O.)
Final correct: Moles H₂O = 1.335 mol → 2.67 mol NaHCO₃ decomposed
Mass decomposed = 2.67 × 84.01 ≈ 224.3 g
Remaining = 100 g − 224.3 g → None remains → contradiction implies limiting reactant stops early
Therefore: Max decomposed = 100 g → 1.19 mol = 0.595 mol H₂O
Mass decomposed = 2.67 × 84.01 ≈ 224.3 g
Remaining = 100 g − 224.3 g → None remains → contradiction implies limiting reactant stops early
Therefore: Max decomposed = 100 g → 1.19 mol = 0.595 mol H₂O
(c) Expression and Value of Kₚ
Kₚ expression:
Kₚ = P_H₂O × P_CO₂ = (x)(x) = x²
From (a), x = P_total / 2 = 7.76 / 2 = 3.88 atm
Kₚ = (3.88)² = 15.05 atm²
✔ Kₚ = 15.05 atm²
(d) Predict Pressure for 110 g of NaHCO₃
Moles of NaHCO₃ = 110 / 84.01 ≈ 1.31 mol → Produces 0.655 mol of H₂O and 0.655 mol of CO₂
Total gas moles = 1.31 mol → P = nRT / V = (1.31 × 0.0821 × 433) / 5.00 ≈ 9.32 atm
✔ Predicted total pressure at equilibrium = 9.32 atm
Because all solid is consumed, the reaction can proceed to completion until this pressure.
Because all solid is consumed, the reaction can proceed to completion until this pressure.