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.