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)
(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

Answer:

A) -1.24 × 10^3 kJ/Mol

Explanation:

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;

Step 1 : Moles of ethanol

We know, Moles = Mass ÷ molar mass

Thus, moles of ethanol = 35.6 g ÷ 46.07 g/mol

                                      = 0.773 moles

Step 2: Enthalpy change or heat change for the reaction.

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