<h3>
Answer:</h3>
A) -1.24 × 10^3 kJ/Mol
<h3>
Explanation:</h3>
we are given;
Mass of ethanol, m = 35.6 g
Temperature change, Δt(35.0 to 76.0°C) = 41 °C
Specific heat capacity of the calorimeter, c = 23.3 kJ/°C
Molar mass of ethanol = 46.07 g/mol
We are required to the heat change of the reaction.
- We need to note that the reaction is an exothermic reaction since there is an increase in temperature which means heat was lost to the surroundings.
Therefore; we are going to use the following steps;
<h3>Step 1 : Moles of ethanol </h3>
We know, Moles = Mass ÷ molar mass
Thus, moles of ethanol = 35.6 g ÷ 46.07 g/mol
= 0.773 moles
<h3>Step 2: Enthalpy change or heat change for the reaction.</h3>
Heat change = -mcΔt
but we are given s[pecific heat capacity in Kj/°C and we require heat change in kJ/mol
Therefore;
Heat change = -(cΔt) ÷ n ( n is the number of moles)
= -( 23.3 kJ/°C × 41°C) ÷0.773 mol
= - 1.24 × 10^3 kJ/Mol
Therefore, values of ΔH of the reaction is -1.24 × 10^3 kJ/Mol
molecule.
molecule.