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) ΔH°rxn = ? (Points : 1)
-1.24 × 103 kJ/mol
+1.24 × 103 kJ/mol
-8.09 × 103 kJ/mol
-9.55 × 103 kJ/mol
+9.55 × 103 kJ/mol

Answer 1

1.24 × 10³ kJ/mol

Step-by-step explanation:

From the question we are given;

Heat capacity of the calorimeter =23.3 kJ/°C

Temperature change, ΔT = 76°C - 35°C

= 41 °C

Mass of ethanol = 35.6 g

Molar mass of ethanol = 46.07 g/mol

We are required to determine the molar enthalpy

We can use the following steps:

Step 1 : Calculate the heat change of the reaction

Heat change will be equivalent to heat gained by the calorimeter.

Therefore;

Heat = heat capacity × change in temperature

Q = CΔT

= 23.33 kJ/°C × 41°C

= 955.3 kJ

Step 2 : Calculate the moles of ethanol burned

Moles = mass ÷ Molar mass

Therefore;

Moles of ethanol = 35.6 g ÷ 46.07 g/mol

= 0.773 moles

Step 3: Calculate the molar enthalpy of the reaction

Heat change for 0.773 moles of ethanol is 955.3 kJ

0.773 moles = 955.3 kJ

1 mole will have ,

= 955.3 kJ ÷ 0.773 moles

= 1235.83 kJ/mol

= 1.24 × 10³ kJ/mol

But since the reaction is exothermic (release of heat) then the enthalpy change will have a negative sign.

Thus;

ΔH = -1.24 × 10³ kJ/mol