Determine the enthalpy change for 7.1 g of C6H6 in the following reaction using the enthalpies of formation. (5 marks) 2C6H6 (l) + 15O2(g) 12CO2(g) + 6H2O(g)

To determine the enthalpy change for the reaction, we need to use the formula:

ΔH = Σ ΔHf (products) - Σ ΔHf (reactants)

First, we need to know the enthalpies of formation (ΔHf) for all the substances involved in the reaction. These values are usually given in the problem or can be found in a table in a chemistry textbook or online. For this problem, let's assume the following values:

ΔHf [C6H6 (l)] = 49 kJ/mol ΔHf [O2 (g)] = 0 kJ/mol (O2 is in its standard state) ΔHf [CO2 (g)] = -394 kJ/mol ΔHf [H2O (g)] = -242 kJ/mol

Next, we multiply the ΔHf values by the stoichiometric coefficients from the balanced chemical equation and add them up for the products and the reactants:

ΔH (products) = 12(-394 kJ/mol) + 6(-242 kJ/mol) = -7080 kJ ΔH (reactants) = 2(49 kJ/mol) + 15(0 kJ/mol) = 98 kJ

Then we substitute these values into the formula:

ΔH = -7080 kJ - 98 kJ = -7178 kJ

This is the enthalpy change for the reaction as written, which involves 2 moles of C6H6. To find the enthalpy change for 7.1 g of C6H6, we need to convert grams to moles using the molar mass of C6H6 (78.11 g/mol):

7.1 g C6H6 * (1 mol/78.11 g) = 0.091 mol C6H6

Finally, we divide the total enthalpy change by the number of moles in the balanced equation (2 moles of C6H6) and multiply by the number of moles we have (0.091 mol):

ΔH (7.1 g C6H6) = (-7178 kJ/2 mol) * 0.091 mol = -327.5 kJ

So, the enthalpy change for 7.1 g of C6H6 in the reaction is -327.5 kJ.