Benzene burns in oxygen according to the following reactions:<br><br>If the standard enthalpies of f — Thermodynamics and Thermochemistry Chemistry Question
Question
Benzene burns in oxygen according to the following reactions:<br>$\text{C}_6\text{H}_6\text{(l)} + \frac{15}{2}\text{O}_2\text{(g)} \to 3\text{H}_2\text{O(l)} + 6\text{CO}_2\text{(g)}$<br>If the standard enthalpies of formation of $\text{C}_6\text{H}_6\text{(l)}$, $\text{H}_2\text{O(l)}$ and $\text{CO}_2\text{(g)}$ are 11.7, –68.1 and –94 kcal/mole, respectively, the amount of heat that will liberate by burning 780 g of benzene is
💡 Solution & Explanation
$\Delta H_c(\text{benzene}) = 6\Delta H_f(\text{CO}_2) + 3\Delta H_f(\text{H}_2\text{O}) - \Delta H_f(\text{C}_6\text{H}_6) = 6(-94) + 3(-68.1) - 11.7 = -564 - 204.3 - 11.7 = -780\text{ kcal/mol}$. Molar mass of benzene is $78\text{ g/mol}$. Moles in 780 g = $10\text{ moles}$. Heat liberated = $10 \times 780\text{ kcal} = 7800\text{ kcal}$.