Question

The heat of combustion of liquid ethanol is $ -327 $ kcal. Calculate the heat of formation of ethanol, given that the heats of formation of $ C{{O}_{2}}(g) $ and $ {{H}_{2}}O(l) $ are $ -94 $ kcal and $ -68.4 $ kcal respectively.
(A) $ 36.2kcal $
(B) $ -36.2kcal $
(C) $ 66.2kcal $
(D) $ -66.2kcal $

Answer 1

Hint: The standard enthalpy of formation (standard heat of formation) of a compound is the change of enthalpy during the formation of one mole of the substance from its constituent elements with all the substances in their standard states.

Complete step by step solution:
You might be thinking that what is the standard state which is necessary for the compounds while calculating the standard enthalpy of formation. Let's talk about it with an example: For example, the standard enthalpy of formation of carbon dioxide gas would be the enthalpy of the following reaction under the standard conditions:
 $ C(graphite)+{{O}_{2}}(g)\to C{{O}_{2}}(g) $
All elements are written in their standard states, and only one mole of product is formed. This is true for all kinds of enthalpies of formation. The above question is based on Hess's law which states that the formation reaction may be considered as the sum of several simpler reactions. In other words, the sum of the enthalpy changes for several individual reactions is equal to the enthalpy change of the overall reaction.
This is true because enthalpy is a state function and not a path function, whose value for an overall process depends only on the initial and final states, not on any intermediate states. Now, by applying Hess's law in the above equation, we get the following chemical equations.
Heat of formation of ethanol $ =\text{ }2\times ~\left[ hea{{t}_{F}}\left( C{{O}_{2}} \right)~ \right]+\text{ }3\times \left[ ~hea{{t}_{F}}\left( {{H}_{2}}O \right)\text{ } \right]-\left[ \text{ }hea{{t}_{C}}\left( ethanol \right) \right] $
 Heat of formation of ethanol $ =~2\times \left( -94 \right)+3\times \left[ \left( -68.4 \right)-\left( -327 \right) \right]~\text{ }=~-66.2kcal $
Therefore, the correct answer is option D i.e. $ -66.2kcal $ .

Note:
Remember that all elements in their standard states (oxygen gas, solid carbon in the form of graphite etc.) have a standard enthalpy of formation of zero, as there is no change in the energy involved in their formation.