First, we need to write the balanced chemical equation for the combustion of ethylene glycol:$$C_2H_6O_2 + \frac{3}3}{2}2}O_2 \rightarrow 2CO_2 + 3H_2O$$Next, we will use Hess's Law to calculate the standard enthalpy change for this reaction. Hess's Law states that the enthalpy change for a reaction is equal to the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants:$$\Delta H_{rxn} = \sum \Delta H_{f, products} - \sum \Delta H_{f, reactants}$$For this reaction, the enthalpies of formation are given as:$$\Delta H_{f, CO_2} = -393.5 \,\text{kJ/mol}$$$$\Delta H_{f, H_2O} = -285.8 \,\text{kJ/mol}$$$$\Delta H_{f, C_2H_6O_2} = -694.3 \,\text{kJ/mol}$$Now, we can plug these values into Hess's Law:$$\Delta H_{rxn} = [2 -393.5 + 3 -285.8 ] - [ -694.3 ]$$$$\Delta H_{rxn} = -787 - 857.4 - -694.3 $$$$\Delta H_{rxn} = -1644.4 + 694.3$$$$\Delta H_{rxn} = -950.1 \,\text{kJ/mol}$$Now that we have the standard enthalpy change for the reaction, we can calculate the enthalpy change for burning 500g of ethylene glycol. First, we need to find the number of moles of ethylene glycol in 500g:The molar mass of ethylene glycol is:$$C_2H_6O_2 = 2 12.01 + 6 1.01 + 2 16.00 = 62.07 \,\text{g/mol}$$So, the number of moles in 500g is:$$\frac{500 \,\text{g}g}}{62.07 \,\text{g/mol}} = 8.05 \,\text{mol}$$Finally, we can calculate the enthalpy change for burning 500g of ethylene glycol:$$\Delta H = -950.1 \,\text{kJ/mol} \times 8.05 \,\text{mol} = -7645.3 \,\text{kJ}$$Therefore, the standard enthalpy change for the combustion of 500g of ethylene glycol is -7645.3 kJ.