Enthalpy of the reaction: is –84.54 kJ. Magnitude of enthalpies of formation of and is in 8:9 ratio — Thermodynamics and Thermochemistry Chemistry Question
Question
Enthalpy of the reaction: $\text{Ag}^+\text{(aq)} + \text{Br}^-\text{(aq)} \to \text{AgBr(s)}$ is –84.54 kJ. Magnitude of enthalpies of formation of $\text{Ag}^+\text{(aq)}$ and $\text{Br}^-\text{(aq)}$ is in 8:9 ratio but their signs are opposite. Enthalpy of formation of AgBr is –99.54 kJ/mol. The magnitude of enthalpy of formation of $\text{Ag}^+\text{(aq)}$ (in kJ/mol) is
💡 Solution & Explanation
$\Delta H_{\text{rxn}} = \Delta H_f(\text{AgBr}) - [\Delta H_f(\text{Ag}^+) + \Delta H_f(\text{Br}^-)] \implies -84.54 = -99.54 - [\Delta H_f(\text{Ag}^+) + \Delta H_f(\text{Br}^-)]$. Thus, $\Delta H_f(\text{Ag}^+) + \Delta H_f(\text{Br}^-) = -15.0\text{ kJ/mol}$. With an 8:9 magnitude ratio and opposite signs, testing $+8x$ and $-9x$ yields $-x = -15 \implies x = 15$. $\Delta H_f(\text{Ag}^+) = 8 \times 15 = 120\text{ kJ/mol}$. (Testing $-8x$ and $+9x$ yields a negative magnitude, which is impossible). The magnitude is 120. Format as 0120.