Question

2. A bomb calorimeter and its contents has a total heat capacity of 17.2 kJ/°C. When 5.00 grams of liquid ethanol, C2H6O (l), is completely combusted in the presence of excess oxygen gas, the calorimeter and its contents increases in temperature by 8.66°C.

a) Write a balanced chemical equation for the complete combustion of one mole of liquid ethanol at 25°C.

b) Determine the enthalpy of combustion (∆H°comb) of liquid ethanol at 25°C in units of kJ/mol.

c) Incomplete combustion of a hydrocarbon can occur when there is not enough oxygen gas to support complete combustion. Incomplete combustion is undesirable because it results in the formation of the toxic product carbon monoxide as well as solid carbon (soot), and it gives off less energy than complete combustion. Using your answer to part (a), along with the following data at 25°C:

2 CO (g) + O2 (g) → 2 CO2 (g) . ∆H° = –566.0 kJ/mol

C (s) + O2 (g) → CO2 (g) ∆H° = –393.5 kJ/mol

Calculate ∆H° for the following reaction at 25°C, which is one reaction that can occur during the incomplete combustion of liquid ethanol:

C2H6O (l) + 3 2 O2 (g) → CO (g) + C (s) + 3 H2O (l) ∆H° = ???

Answer 1

(a) Balanced equation for the complete combustion of one mole of liquid ethanol at 25°C. :

C2H5OH +3O2 → 2CO2 + 3H2O……… (1)

(b) Total heat released in combustion of 5 g C2H5OH= total heat capacity of bomb calorimeter and its contents x change in temperature

= 17.2 Kj/°C x 8.66°C.

= 148.952 kj

Mol of C2H5OH=( gram of C2H5OH/ molecular weight of C2H5OH)

=5/46

=0.1087

So, molar heat of combustion =-148.952 kj / Mol of C2H5OH

=-(148.952 kj/0.1087 mol)

=-1370.30 kj/mol

(c) For incomplete combustion the equation is

C2H6O (l) + 3/2 O2 (g) → CO (g) + C (s) + 3 H2O (l) …………(4)

The given equations and their ∆H° values are

2 CO (g) + O2 (g) → 2 CO2 (g) . ∆H° = –566.0 kJ/mol ………… (2)

C (s) + O2 (g) → CO2 (g) ∆H° = –393.5 kJ/mol ……….. (3)

Equation (4) can be represented by rearranging equation (1), (2) and (3) as follows

Equation (1) - (1/2) x Equation (2) – Equation (3)

C2H5OH(l) +3O2 (g) →    2CO2 (g) + 3H2O (l)

-(1/2) {2 CO (g) + O2 (g) → 2 CO2}

- {C (s) + O2 (g) → CO2}

C2H6O (l) + 3/2 O2 (g) → CO (g) + C (s) + 3 H2O (l)

So, ∆H° for incomplete combustion = -1370.35 - (1/2)x(–566.0)-(–393.5) kJ/mol

= – 693.85 kJ/mol