For combustion of one mole of magnesium in an open container at 300 K and 1 bar pressure, ∆H = –601. — Thermodynamics and Thermochemistry Chemistry Question
Question
For combustion of one mole of magnesium in an open container at 300 K and 1 bar pressure, ∆H = –601.70 kJ mol , the magnitude of change in internal energy for the reaction is ____ kJ. (Nearest integer) (Given: R = 8.3 J K mol ) C ⊖ –1 –1 –1
💡 Solution & Explanation
**Step 1: Identify the relationship between ∆H and ∆U** For a reaction at constant temperature and pressure: $$\Delta H = \Delta U + \Delta(PV) = \Delta U + \Delta n_g RT$$ where ∆n_g = change in moles of gaseous products minus gaseous reactants **Step 2: Write the balanced combustion equation** $$2Mg(s) + O_2(g) \rightarrow 2MgO(s)$$ **Step 3: Calculate ∆n_g for one mole of Mg** From the equation: 2 mol Mg requires 1 mol O₂, producing 0 mol gases (MgO is solid) For 1 mole of Mg: $$\Delta n_g = 0 - \frac{1}{2} = -0.5 \text{ mol}$$ **Step 4: Calculate ∆(n_g RT)** $$\Delta n_g RT = (-0.5) \times 8.3 \times 300$$ $$= -0.5 \times 2490 = -1245 \text{ J} = -1.245 \text{ kJ}$$ **Step 5: Calculate ∆U** $$\Delta U = \Delta H - \Delta n_g RT$$ $$\Delta U = -601.70 - (-1.245)$$ $$\Delta U = -601.70 + 1.245 = -600.455 \text{ kJ}$$ **Step 6: Find the magnitude** $$|\Delta U| = 600.455 \approx 600 \text{ kJ}$$ Therefore, the answer is **600**.