A35.6 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:

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.

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.