Question

A student performed an experiment to measure the enthalpy change of combustion of ethane.
He used the following values for the standard enthalpy changes of combustion of carbon and hydrogen.
carbon = –394 kJ mol–1 hydrogen = –286 kJ mol–1
He calculated the enthalpy change of formation of ethane to be –140 kJ mol–1.
What was his experimental value for the standard enthalpy change of combustion of ethane?

Answer 1

Answer on Question #60100 – Chemistry | Inorganic Chemistry

Calculating the standard enthalpy of combustion

Solution:

1) Hess' Law:

$$\Delta H ^ {\circ} _ {\text {r x n}} = \Sigma \Delta H ^ {\circ} _ {\text {f , p r o d u c t s}} - \Sigma \Delta H ^ {\circ} _ {\text {f , r e a c t a n t s}}$$

The standard formation reaction for the ethane, carbon dioxide, and water:

$$2 C (s) + 3 H _ {2} (g) = C _ {2} H _ {6} (g) \quad \Delta H ^ {\circ} _ {f, C 2 H 6}: - 1 4 0 k J ^ {*} m o l ^ {- 1}$$ $$C (s, g r.) + O _ {2} (g) = C O _ {2} (g) \quad \Delta H ^ {\circ} _ {f, C O 2}: - 3 9 4 k J ^ {*} m o l ^ {- 1}$$ $$\mathrm {H} _ {2} (\mathrm {g}) + 1 / 2 \mathrm {O} _ {2} (\mathrm {g}) = \mathrm {H} _ {2} \mathrm {O} (\mathrm {l}) \quad \Delta \mathrm {H} ^ {\circ} _ {\mathrm {f}, \mathrm {H} 2 \mathrm {O}}: - 2 8 6 \mathrm {k J} ^ {*} \mathrm {m o l} ^ {- 1}$$

Meaning that the standard enthalpy changes of formation of water and carbon dioxide are exactly the same as the enthalpy change of combustion of Carbon and Hydrogen.

2) Substitute values into the equation. Since oxygen is an element in its standard state, its enthalpy of formation is zero.

$$\Delta H ^ {\circ} _ {f, O 2}: 0$$

The equation for the combustion of $\mathrm{C_2H_6}$ (ethane) is:

$$\mathrm {C} _ {2} \mathrm {H} _ {6} + 7 / 2 \mathrm {O} _ {2} = 2 \mathrm {C O} _ {2} + 3 \mathrm {H} _ {2} \mathrm {O}$$ $$[ 2 ^ {*} (- 3 9 4) + 3 ^ {*} (- 2 8 6) ] - [ (- 1 4 0) + 7 / 2 ^ {*} 0 ] = - 1 5 0 6 k J ^ {*} m o l ^ {- 1}$$

http://www.AssignmentExpert.com/