The enthalpies of combustion of formaldehyde and paraformaldehyde (a polymer of formaldehyde) are –1 — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Let the polymer be $(\text{CH}_2\text{O})_n$. The reaction is $n \text{CH}_2\text{O} \to (\text{CH}_2\text{O})_n$ with $\Delta H = -72\text{ kcal}$. Using combustion data: $\Delta H_c(\text{polymer}) = n \Delta H_c(\text{monomer}) - \Delta H_{\text{poly}} \implies -732 = n(-134) - (-72) \implies -732 = -134n + 72 \implies 134n = 804 \implies n = 6$. The formula is $(\text{CH}_2\text{O})_6 = \text{C}_6\text{H}_{12}\text{O}_6$.