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Chapter 5: Problem 37

The complete combustion of acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH}(l)\), to form \(\mathrm{H}_{2} \mathrm{O}(l)\) and \(\mathrm{CO}_{2}(g)\) at constant pressure releases \(871.7 \mathrm{~kJ}\) of heat per mole of \(\mathrm{CH}_{3} \mathrm{COOH}\). (a) Write a balanced thermochemical equation for this reaction. (b) Draw an enthalpy diagram for the reaction.

Short Answer

Expert verified
The balanced thermochemical equation for the complete combustion of acetic acid is: \(CH_3COOH(l) + 2O_2(g) 鈫� 2H_2O(l) + 2CO_2(g)\), with 螖H = -871.7 kJ/mol. For the enthalpy diagram, start with a horizontal line representing the reactants (CH3COOH(l) + 2O2(g)), and another horizontal line representing the products (2H2O(l) + 2CO2(g)) slightly below the reactants line. Connect these lines with a diagonal arrow pointing downwards and to the right, labeled as "螖H = -871.7 kJ/mol". This shows that energy is released during the combustion of acetic acid.

Step by step solution

01

(a) Write a balanced thermochemical equation for the reaction

First, we need to write the unbalanced chemical equation for the complete combustion of acetic acid: CH3COOH(l) + O2(g) --> H2O(l) + CO2(g) Next, we need to balance the equation. We can start by balancing the carbon atoms: CH3COOH(l) + O2(g) --> H2O(l) + 2CO2(g) Now, let's balance the hydrogen atoms: CH3COOH(l) + O2(g) --> 2H2O(l) + 2CO2(g) Finally, we can balance the oxygen atoms: CH3COOH(l) + 2O2(g) --> 2H2O(l) + 2CO2(g) Now that we have a balanced equation, we can incorporate the heat released per mole of CH3COOH: CH3COOH(l) + 2O2(g) 鈫� 2H2O(l) + 2CO2(g) 螖H = -871.7 kJ/mol
02

(b) Draw an enthalpy diagram for the reaction

To draw an enthalpy diagram for the reaction, we'll begin by labeling the reactants and products and indicating the change in enthalpy (螖H) for the reaction: 1. Draw a horizontal line on the left side and label it as "Reactants: CH3COOH(l) + 2O2(g)". 2. Draw a horizontal line on the right side, slightly lower than the reactants line, and label it as "Products: 2H2O(l) + 2CO2(g)". 3. Draw a diagonal arrow pointing down and right from the reactants line to the products line. Label the arrow as "螖H = -871.7 kJ/mol". The enthalpy diagram should show that the products have lower energy than the reactants, indicating that energy is released during the combustion of acetic acid.

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Most popular questions from this chapter

(a) Why are tables of standard enthalpies of formation so useful? (b) What is the value of the standard enthalpy of formation of an element in its most stable form? (c) Write the chemical equation for the reaction whose enthalpy change is the standard enthalpy of formation of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{~s}), \Delta \mathrm{H}_{f}^{\circ}\left[\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right]\).

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