A 35.6 g sample of ethanol (C2H5OH) is burned in a bomb calorimeter, according to the following reaction. If the temperature rose from 35.0 to 76.0 C and the heat capacity of the calorimeter is 23.3 kJ / C, what is the value of DH rxn? The molar mass of ethanol is 46.07 g/mol.

C2H5OH (l) + O2 (g) - -> CO2 (g) + H2O (g) ? H_ rxn = ?

A) - 1.24 * 103 kJ/mol

B) + 1.24 * 103 kJ/mol

C) - 8.09 * 103 kJ/mol

D) - 9.55 * 103 kJ/mol

E) + 9.55 * 103 kJ/mol

A) - 1.24 * 10³ kJ/mol

Explanation:

To solve this problem we can use two formulas:

Q = C*ΔT ΔH = Q/mol

So first let's calculate Q, using the heat capacity C and the difference in temperature ΔT:

Q = - 23.3 kJ/°C * (76.0-35-0) °C

Q = 955.4 kJ

Then let's calculate the moles of ethanol, using its molar mass:

mol ethanol = 35.6 g : 46.07 g/mol = 0.773 mol

Finally we calculate ΔH:

ΔH = - 955.4kJ/0.773mol

ΔH = - 1235.96 kJ/mol ≅ - 1.24 * 10³ kJ/mol

It has a negative value because it is an exothermic reaction.