Question

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

Answer 1

A) -1.24 * 10³ kJ/mol

Step-by-step explanation:

To solve this problem we can use two formulas:

1. Q = C*ΔT
2. 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.