Question

Calculate the enthalpy of formation of ethanoic acid from the following data. a) Enthalpy of formation of carbon dioxide is $393.5$ $KJmo{l^{ - 1}}$. b) The enthalpy of formation of water is $- 285.8$ $KJmo{l^{ - 1}}$. Enthalpy of combustion of ethanoic acid is - 875$$$$KJmo{l^{ - 1}}.

Answer 1

We can write the equation for the combustion of ethanoic acid as $C{H_3}COOH + 3{O_2}(g) \to 2C{O_2}(g) + 2{H_2}O(l)$. Where, enthalpy of formation of carbon dioxide, the enthalpy of formation of water and the enthalpy of combustion of ethanoic acid is given in the question as -393.5$$$$KJmo{l^{ - 1}}, - 285.8$$$$KJmo{l^{ - 1}}and - 875$$$$KJmo{l^{ - 1}}respectively. Using the equation $\Delta {H^0} = \sum \Delta {H^0}_f(product) - \sum \Delta {H^0}_f(reactant)$, we can calculate the enthalpy of formation of ethanoic acid.

Complete step by step answer:
Given in the question are:
Enthalpy of formation of carbon dioxide = - 393.5$$$$KJmo{l^{ - 1}}
The enthalpy of formation of water = - 285.8$$$$KJmo{l^{ - 1}}
Enthalpy of combustion of ethanoic acid $= - 875$ $KJmo{l^{ - 1}}$
Let enthalpy of formation of acetic acid be x kJ/mol.
The combustion of ethanoic acid-
$C{H_3}COOH + 3{O_2}(g) \to 2C{O_2}(g) + 2{H_2}O(l)\Delta H = - 875kJ/mol$
$\Delta {H^0} = \sum \Delta {H^0}_f(product) - \sum \Delta {H^0}_f(reactant)$
$\therefore - 875 = (2 \times ( - 393.5) + 2 \times ( - 285.8)) - (x + 0)$
$\Rightarrow x = - 1358.6 + 867 = - 491.6kJ/mol$

Hence the enthalpy of formation of acetic acid will be $-491.6$kcal/mol.

Note: The standard conditions for which most thermochemical data are tabulated are a pressure of 1 atmosphere (atm) for all gases and a concentration of 1 M for all species in solution (1 mol/L). In addition, each pure substance must be in its standard state, which is usually its most stable form at a pressure of 1 atm at a specified temperature. We assume a temperature of $25^\circ C$ (298 K) for all enthalpy changes given in this text, unless otherwise indicated. Enthalpies of formation measured under these above conditions are called standard enthalpies of formation ($\Delta {H^0}_f$) The enthalpy change for the formation of 1 mol of a compound from its component elements when the component elements are each in their standard states. The standard enthalpy of formation of any element in its most stable form is zero by definition.