A35.6 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

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

10.063g / 10.00 ml = 1.0063 g/ml which rounds off to 4 sig figs as: 1.006 g/ml

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