200 mL of 0.2 M HCl is mixed with 300 mL of 0.1 M NaOH. The molar heat of neutralization of this rea — Thermodynamics and Thermochemistry Chemistry Question
Question
200 mL of 0.2 M HCl is mixed with 300 mL of 0.1 M NaOH. The molar heat of neutralization of this reaction is –57.1 kJ. The increase in temperature in °C of the system on mixing is x × 10 . The value of x is ________ . (Nearest integer) [Given: Specific heat of water = 4.18 J g K Density of water = 1.00 g cm ] (Assume no volume change on mixing) : –2 –1 –1 –3
💡 Solution & Explanation
**Step 1: Calculate moles of HCl and NaOH** - Moles of HCl = 0.2 M × 0.2 L = 0.04 mol - Moles of NaOH = 0.1 M × 0.3 L = 0.03 mol **Step 2: Identify the limiting reagent** HCl is in excess. NaOH is the limiting reagent. - Moles of HCl that react = 0.03 mol **Step 3: Calculate heat released** Using: q = moles × molar heat of neutralization - q = 0.03 mol × (–57.1 kJ/mol) = –1.713 kJ = –1713 J - Heat released = 1713 J (absolute value) **Step 4: Calculate total mass of solution** Assuming density = 1.00 g/mL and no volume change: - Total volume = 200 + 300 = 500 mL - Total mass = 500 g **Step 5: Calculate temperature change** Using: q = m × c × ΔT - 1713 = 500 × 4.18 × ΔT - ΔT = 1713/(500 × 4.18) = 1713/2090 = 0.8197 K ≈ 0.82°C **Step 6: Express in the form x × 10⁻²** - ΔT = 0.82°C = 82 × 10⁻² °C - Therefore x = 82 Therefore, the answer is 82.00.