When 600 mL of 0.2 M HNO is mixed with 400 mL of 0.1M NaOH solution in a flask, the rise in temperat — Thermodynamics and Thermochemistry Chemistry Question
Question
When 600 mL of 0.2 M HNO is mixed with 400 mL of 0.1M NaOH solution in a flask, the rise in temperature of the flask is _______ × 10 C. (Enthalpy of neutralisation = 57 kJ mol and Specific heat of water = 4.2 JK g ) (Neglect heat capacity of flask) 3 –2 o –1 –1 –1
💡 Solution & Explanation
**Step 1: Find the limiting reagent** Moles of HNO₃ = 0.6 L × 0.2 M = 0.12 mol Moles of NaOH = 0.4 L × 0.1 M = 0.04 mol NaOH is the limiting reagent (smaller amount). **Step 2: Calculate heat released** The neutralisation reaction: HNO₃ + NaOH → NaNO₃ + H₂O Heat released = moles of limiting reagent × enthalpy of neutralisation Q = 0.04 mol × 57 kJ/mol = 2.28 kJ = 2280 J **Step 3: Find total mass of solution** Total volume = 600 + 400 = 1000 mL = 1000 g (Assuming density of solution ≈ 1 g/mL) **Step 4: Apply heat equation** Q = m × c × ΔT Where: Q = 2280 J, m = 1000 g, c = 4.2 J·K⁻¹·g⁻¹ **Step 5: Solve for temperature rise** ΔT = Q / (m × c) ΔT = 2280 / (1000 × 4.2) ΔT = 2280 / 4200 ΔT = 0.543°C ≈ 0.54°C = 54 × 10⁻² °C Therefore, the answer is **54** (in units of 10⁻² °C).