<h3>
Answer:</h3>
1.24 × 10³ kJ/mol
<h3>
Explanation:</h3>
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:
<h3> Step 1 : Calculate the heat change of the reaction</h3>
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
<h3>Step 2 : Calculate the moles of ethanol burned </h3>
Moles = mass ÷ Molar mass
Therefore;
Moles of ethanol = 35.6 g ÷ 46.07 g/mol
= 0.773 moles
<h3>Step 3: Calculate the molar enthalpy of the reaction </h3>
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
