The heat of total cracking of hydrocarbons, is defined as at 298.15 K and 101.325 kPa for the proces — Thermodynamics and Thermochemistry Chemistry Question
Question
The heat of total cracking of hydrocarbons, $\Delta H_{\text{TC}}$ is defined as $\Delta H$ at 298.15 K and 101.325 kPa for the process: $\text{C}_n\text{H}_m + \left(2n - \frac{m}{2}\right)\text{H}_2\text{(g)} \to n\text{CH}_4\text{(g)}$. The values of $\Delta H_{\text{TC}}$ is –65.2 kJ for $\text{C}_2\text{H}_6$ and –87.4 kJ for $\text{C}_3\text{H}_8$. Calculate $\Delta H$ (in kJ) for $\text{CH}_4\text{(g)} + \text{C}_3\text{H}_8\text{(g)} \to 2\text{C}_2\text{H}_6\text{(g)}$
💡 Solution & Explanation
Eq 1 (for $\text{C}_2\text{H}_6$): $\text{C}_2\text{H}_6 + \text{H}_2 \to 2\text{CH}_4$ ($\Delta H = -65.2\text{ kJ}$). Eq 2 (for $\text{C}_3\text{H}_8$): $\text{C}_3\text{H}_8 + 2\text{H}_2 \to 3\text{CH}_4$ ($\Delta H = -87.4\text{ kJ}$). Reverse Eq 1 and multiply by 2: $4\text{CH}_4 \to 2\text{C}_2\text{H}_6 + 2\text{H}_2$ ($\Delta H = +130.4\text{ kJ}$). Adding this to Eq 2 yields: $\text{C}_3\text{H}_8 + 4\text{CH}_4 + 2\text{H}_2 \to 3\text{CH}_4 + 2\text{C}_2\text{H}_6 + 2\text{H}_2$, simplifying to $\text{C}_3\text{H}_8 + \text{CH}_4 \to 2\text{C}_2\text{H}_6$. Total $\Delta H = 130.4 - 87.4 = 43.0\text{ kJ}$. Format as 0043.