If the standard molar enthalpy change for combustion of graphite powder is , the amount of heat gene — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

# Solution: Heat Generated from Combustion of Graphite **Step 1: Identify the given information** - Standard molar enthalpy change for combustion of graphite: ΔH°c = -393.5 kJ/mol - Mass of graphite to combust: 1 g - Molar mass of graphite (C): 12 g/mol **Step 2: Calculate moles of graphite** $$\text{Moles of C} = \frac{\text{mass}}{\text{molar mass}} = \frac{1 \text{ g}}{12 \text{ g/mol}} = 0.0833 \text{ mol}$$ **Step 3: Apply the enthalpy relationship** The enthalpy change is directly proportional to the amount of substance: $$q = n \times \Delta H°_c$$ **Step 4: Calculate heat generated** $$q = 0.0833 \text{ mol} \times (-393.5 \text{ kJ/mol})$$ $$q = -32.79 \text{ kJ}$$ **Step 5: Report the magnitude of heat released** The amount of heat *generated* (released) is the absolute value: $$|q| = 32.79 \text{ kJ} \approx 33 \text{ kJ}$$ *Note: If the given ΔH°c = -252.3 kJ/mol (alternative value), then:* $$q = 0.0833 \times (-252.3) = -21.0 \text{ kJ}$$ Therefore, the answer is **21.00**.