Question

Glycine, C2H5O2N, is important for biological energy. The combustion reaction of glycine is described by the following thermochemical equation. 4C2H5O2N(s) + 9O2(g) → 8CO2(g) + 10H2O(l) + 2N2(g) ΔH°rxn = –3896 kJ/molWhat is the standard enthalpy of formation of solid glycine?–51.90 kJ/mol–527.5 kJ/mol–974.0 kJ/mol–1502 kJ/mol–2476 kJ/mol

Answer 1

Answer:

-537.25 kJ/mol is the standard enthalpy of formation of solid glycine.

Explanation:

$4C_2H_5O_2N(s) + 9O_2(g)\rightarrow 8CO_2(g) + 10H_2O(l) + 2N_2(g) ,\Delta H^{rxn} =-3896 kJ/mol$

Standard enthalpy of formation of oxygen gas= $\Delta H_{f,O_2}=0$

Standard enthalpy of formation of carbon dioxide= $\Delta H_{f,CO_2}=-393.5 kJ/mol$

Standard enthalpy of formation of water = $\Delta H_{f,H_2O}=-285.8 kJ/mol$

Standard enthalpy of formation of nitrogen gas= $\Delta H_{f,N_2}=0$

Standard enthalpy of formation of glycine = $\Delta H_{f,gly}=?$

Enthalpy of the reaction = $H_{rxn}=-3896 kJ/mol$

$H_{rxn} =$

=$8\times \Delta H_{f,CO_2} +10\times \Delta H_{f,H_2O}+2\times \Delta H_{f,N_2} - (4\times \Delta H_{f,gly}+9\times \Delta H_{f,O_2})$

$-3896 kJ/mol=8\times (-393.5 kJ/mol)+ 10\times (-285.8 kJ/mol)+0 - (4\times \Delta H_{f,gly} +0)$

On rearranging :

$4\times \Delta H_{f,gly}=8\times (-393.5 kJ/mol)+ 10\times (-285.8 kJ/mol)+ 3896 kJ/mol$

$\Delta H_{f,gly}=\frac{-2149 kJ/mol}{4}=-537.25 kJ/mol$

-537.25 kJ/mol is the standard enthalpy of formation of solid glycine.