Question

Calculate the heat of formation of acetic acid from the following data:

$${\text{a}}{\text{. C}}\left( {\text{s}} \right) + {{\text{O}}_2}\left( {\text{g}} \right) \to {\text{C}}{{\text{O}}_2}\left( {\text{g}} \right),{\text{ }}\Delta {\text{H}} = - 393.7{\text{ KJ}} \\\ {\text{b}}{\text{. }}{{\text{H}}_2}\left( {\text{g}} \right) + \dfrac{1}{2}{{\text{O}}_2}\left( {\text{g}} \right) \to {{\text{H}}_2}{\text{O}}\left( {\text{l}} \right),{\text{ }}\Delta {\text{H}} = - 285.8{\text{ KJ}} \\\ {\text{c}}{\text{. C}}{{\text{H}}_3}{\text{COOH}}\left( {\text{l}} \right) + 2{{\text{O}}_2}\left( {\text{g}} \right) \to 2{\text{C}}{{\text{O}}_2}\left( {\text{g}} \right) + 2{{\text{H}}_2}{\text{O}}\left( {\text{l}} \right),{\text{ }}\Delta {\text{H}} = - 873.2{\text{ KJ}} \\\$$ $${\text{A}}{\text{. }}{\Delta _f}{\text{H}} = - 485.8{\text{ KJmo}}{{\text{l}}^{ - 1}} \\\ {\text{B}}{\text{. }}{\Delta _f}{\text{H}} = + 485.8{\text{ KJmo}}{{\text{l}}^{ - 1}} \\\ {\text{C}}{\text{. }}{\Delta _f}{\text{H}} = - 1432.2{\text{ KJmo}}{{\text{l}}^{ - 1}} \\\$$

${\text{D}}{\text{.}}$ None of these

Answer 1

Hint: Here, we will proceed by reversing the third given chemical equation in order to get the acetic acid in the product side and due to this, the corresponding sign of enthalpy is also reversed. Finally, we will perform some operation to obtain the chemical reaction of formation of acetic acid with the help of the given chemical equations.
Complete answer:
We are given with various chemical reactions and corresponding to these, their enthalpies are also given.
The first chemical reaction is given as ${\text{C}}\left( {\text{s}} \right) + {{\text{O}}_2}\left( {\text{g}} \right) \to {\text{C}}{{\text{O}}_2}\left( {\text{g}} \right){\text{ }} \to {\text{a}}$ and its enthalpy is ${\left( {\Delta {\text{H}}} \right)_{\text{a}}} = - 393.7{\text{ KJ}}$
The second chemical reaction is given as ${{\text{H}}_2}\left( {\text{g}} \right) + \dfrac{1}{2}{{\text{O}}_2}\left( {\text{g}} \right) \to {{\text{H}}_2}{\text{O}}\left( {\text{l}} \right){\text{ }} \to {\text{b}}$ and its enthalpy is ${\left( {\Delta {\text{H}}} \right)_{\text{b}}} = - 285.8{\text{ KJ}}$
The third chemical reaction can be reversed and can be written conveniently as $2{\text{C}}{{\text{O}}_2}\left( {\text{g}} \right) + 2{{\text{H}}_2}{\text{O}}\left( {\text{l}} \right) \to {\text{C}}{{\text{H}}_3}{\text{COOH}}\left( {\text{l}} \right) + 2{{\text{O}}_2}\left( {\text{g}} \right){\text{ }} \to {\text{c}}$ and whenever any chemical reaction is reversed, the sign of enthalpy is changed
i.e., ${\left( {\Delta {\text{H}}} \right)_{\text{c}}} = - \left( { - 873.2} \right){\text{ KJ}} = 873.2{\text{ KJ}}$
As we know that the balanced chemical reaction corresponding to the formation of the acetic acid is given by
$2{\text{C}}\left( {\text{s}} \right) + 2{{\text{H}}_2}\left( {\text{g}} \right) + {{\text{O}}_2}\left( {\text{g}} \right) \to {\text{C}}{{\text{H}}_3}{\text{COOH}}\left( {\text{l}} \right){\text{ }} \to 1$
i.e., when 2 moles of carbon in solid form are mixed with 2 moles of hydrogen in gaseous state and 1 mole of oxygen in gaseous state, acetic acid in liquid state will be formed as the product.
In order to get the chemical reaction given by equation (1) with the help of the chemical reactions given by equations a, b and c, we can perform the below operation
i.e., 2(equation a) + 2(equation b) + (equation c)

$$2\left[ {{\text{C}}\left( {\text{s}} \right) + {{\text{O}}_2}\left( {\text{g}} \right)} \right] + 2\left[ {{{\text{H}}_2}\left( {\text{g}} \right) + \dfrac{1}{2}{{\text{O}}_2}\left( {\text{g}} \right)} \right] + 2{\text{C}}{{\text{O}}_2}\left( {\text{g}} \right) + 2{{\text{H}}_2}{\text{O}}\left( {\text{l}} \right) \to 2\left[ {{\text{C}}{{\text{O}}_2}\left( {\text{g}} \right)} \right] + 2\left[ {{{\text{H}}_2}{\text{O}}\left( {\text{l}} \right)} \right] + {\text{C}}{{\text{H}}_3}{\text{COOH}}\left( {\text{l}} \right) + 2{{\text{O}}_2}\left( {\text{g}} \right) \\\ {\text{2C}}\left( {\text{s}} \right) + 2{{\text{O}}_2}\left( {\text{g}} \right) + {\text{2}}{{\text{H}}_2}\left( {\text{g}} \right) + {{\text{O}}_2}\left( {\text{g}} \right) + 2{\text{C}}{{\text{O}}_2}\left( {\text{g}} \right) + 2{{\text{H}}_2}{\text{O}}\left( {\text{l}} \right) \to {\text{2C}}{{\text{O}}_2}\left( {\text{g}} \right) + {\text{2}}{{\text{H}}_2}{\text{O}}\left( {\text{l}} \right) + {\text{C}}{{\text{H}}_3}{\text{COOH}}\left( {\text{l}} \right) + 2{{\text{O}}_2}\left( {\text{g}} \right) \\\ {\text{2C}}\left( {\text{s}} \right) + {\text{2}}{{\text{H}}_2}\left( {\text{g}} \right) + \left[ {2{{\text{O}}_2}\left( {\text{g}} \right) + {{\text{O}}_2}\left( {\text{g}} \right) - 2{{\text{O}}_2}\left( {\text{g}} \right)} \right] + \left[ {2{\text{C}}{{\text{O}}_2}\left( {\text{g}} \right) - {\text{2C}}{{\text{O}}_2}\left( {\text{g}} \right)} \right] + \left[ {2{{\text{H}}_2}{\text{O}}\left( {\text{l}} \right) - {\text{2}}{{\text{H}}_2}{\text{O}}\left( {\text{l}} \right)} \right] \to {\text{C}}{{\text{H}}_3}{\text{COOH}}\left( {\text{l}} \right) \\\ {\text{2C}}\left( {\text{s}} \right) + {\text{2}}{{\text{H}}_2}\left( {\text{g}} \right) + {{\text{O}}_2}\left( {\text{g}} \right) \to {\text{C}}{{\text{H}}_3}{\text{COOH}}\left( {\text{l}} \right) \\\$$

Clearly, the above obtained equation is the same as equation (1).
Since, equation 1 = 2(equation a) + 2(equation b) + (equation c)
So, the heat of formation of the acetic acid will be given by
${\Delta _f}{\text{H}} = 2{\left( {\Delta {\text{H}}} \right)_{\text{a}}} + 2{\left( {\Delta {\text{H}}} \right)_{\text{b}}} + {\left( {\Delta {\text{H}}} \right)_{\text{c}}}$
By substituting the given values in the above equation, we get

$${\Delta _f}{\text{H}} = 2{\left( {\Delta {\text{H}}} \right)_{\text{a}}} + 2{\left( {\Delta {\text{H}}} \right)_{\text{b}}} + {\left( {\Delta {\text{H}}} \right)_{\text{c}}} \\\ \Rightarrow {\Delta _f}{\text{H}} = 2\left( { - 393.7} \right) + 2\left( { - 285.8} \right) + 873.2 \\\ \Rightarrow {\Delta _f}{\text{H}} = - 787.4 - 571.6 + 873.2 \\\ \Rightarrow {\Delta _f}{\text{H}} = - 485.8{\text{ KJmo}}{{\text{l}}^{ - 1}} \\\$$

The units of the enthalpy of formation of acetic acid is written as ${\text{KJmo}}{{\text{l}}^{ - 1}}$ because only 1 mole of acetic acid is obtained.
Therefore, the required heat or enthalpy of formation of acetic acid is $- 485.8{\text{ KJmo}}{{\text{l}}^{ - 1}}$.

Hence, option A is correct.

Note: The amount of the heat absorbed or evolved when one mole of a substance is generated as the product by reacting various constituent elements (reactants) together which are in their standard states (solid, liquid or gaseous) is known as the heat of formation or standard enthalpy of formation.