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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: \(CH_3COOH(l) + 2O_2(g) → 2H_2O(l) + 2CO_2(g)\), with ΔH = -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 "ΔH = -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) → 2H2O(l) + 2CO2(g) ΔH = -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 (ΔH) 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 "ΔH = -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 200-lb man decides to add to his exercise routine by walking up three flights of stairs ( \(45 \mathrm{ft}\) ) 20 times per day. He figures that the work required to increase his potential energy in this way will permit him to eat an extra order of French fries, at 245 Cal, without adding to his weight. Is he correct in this assumption?

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Calculate \(\Delta E\), and determine whether the process is endothermic or exothermic for the following cases: (a) A system absorbs \(105 \mathrm{~kJ}\) of heat from its surroundings while doing \(29 \mathrm{~kJ}\) of work on the surroundings; (b) \(q=1.50 \mathrm{~kJ}\) and \(w=-657 \mathrm{~J} ;\) (c) the system releases \(57.5 \mathrm{~kJ}\) of heat while doing \(22.5 \mathrm{~kJ}\) of work on the surroundings.

(a) When a 0.235-g sample of benzoic acid is combusted in a bomb calorimeter, the temperature rises \(1.642^{\circ} \mathrm{C}\). When a 0.265-g sample of caffeine, \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{O}_{2} \mathrm{~N}_{4}\), is burned, the temperature rises \(1.525^{\circ} \mathrm{C}\). Using the value \(26.38 \mathrm{~kJ} / \mathrm{g}\) for the heat of combustion of benzoic acid, calculate the heat of combustion per mole of caffeine at constant volume. (b) Assuming that there is an uncertainty of \(0.002^{\circ} \mathrm{C}\) in each temperature reading and that the masses of samples are measured to \(0.001 \mathrm{~g}\), what is the estimated uncertainty in the value calculated for the heat of combustion per mole of caffeine?

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