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Chapter 6: Problem 95

Formic acid, \(\mathrm{HCHO}_{2}\), was first discovered in ants ( formica is Latin for "ant"). In an experiment, \(5.48 \mathrm{~g}\) of formic acid was burned at constant pressure. $$ 2 \mathrm{HCHO}_{2}(l)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ If \(30.3 \mathrm{~kJ}\) of heat evolved, what is \(\Delta H\) per mole of formic acid?

Short Answer

Expert verified
\(\Delta H\) is approximately \(-254.6 \text{ kJ/mol}\).

Step by step solution

01

Identify the given information

We have an experimental setup where 5.48 g of formic acid is combusted, releasing 30.3 kJ of energy. The molar mass of formic acid (HCHO鈧�) needs to be calculated first, with hydrogen (H) being 1 g/mol, carbon (C) being 12 g/mol, and oxygen (O) being 16 g/mol.
02

Calculate the molar mass of formic acid

Formic acid, HCHO鈧�, has a molar mass calculated as follows: 1 (H) + 12 (C) + 16*2 (O) = 46 g/mol.
03

Determine moles of formic acid combusted

The moles of formic acid combusted can be found using the formula \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). Thus, moles of formic acid = \( \frac{5.48 \text{ g}}{46 \text{ g/mol}} \approx 0.119 \text{ mol} \).
04

Determine the heat evolved per mole of formic acid

The total heat evolved in the experiment is 30.3 kJ. To find the heat per mole of formic acid, use the formula \( \Delta H = \frac{\text{total heat released}}{\text{moles of formic acid}} \). Thus, \[ \Delta H = \frac{30.3 \text{ kJ}}{0.119 \text{ mol}} \approx 254.6 \text{ kJ/mol} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combustion Reaction
A combustion reaction involves a substance reacting quickly with oxygen to produce heat and light. In this exercise, formic acid undergoes a combustion reaction. The equation given is:
  • 2 \(\mathrm{HCHO}_{2}\)(l) + \(\mathrm{O}_{2}\)(g) \(\longrightarrow\) 2 \(\mathrm{CO}_{2}\)(g) + 2 \(\mathrm{H}_{2}\0\)(l)
This reaction shows that formic acid is oxidized, turning into carbon dioxide and water.
This process releases energy, making it an exothermic reaction.
It's important to note that combustion reactions always need oxygen as a reactant.When dealing with such reactions, the goal is often to calculate energy changes, such as enthalpy changes, which are covered in later sections.
Molar Mass Calculation
Before you can determine the energy changes in a chemical reaction, you need to calculate molar masses. The molar mass is the mass of one mole of a substance. For formic acid, \(\mathrm{HCHO}_{2}\), calculating the molar mass involves adding the atomic masses from the periodic table:
  • Hydrogen: 1 g/mol
  • Carbon: 12 g/mol
  • Oxygen (each): 16 g/mol
Adding these up for \(\mathrm{HCHO}_{2}\):
1 (H) + 12 (C) + 16 \(\times\) 2 (O) = 46 g/mol.
Knowing the molar mass allows you to convert between grams and moles, which is key when calculating heat per mole in chemical reactions.
Enthalpy Change Calculation
Enthalpy change (\(\Delta H\)) measures the change in heat content during a reaction at constant pressure. It's expressed in kilojoules per mole \((\text{kJ/mol})\).
To calculate \(\Delta H\) for a reaction, use the formula:
  • \(\Delta H = \frac{\text{total heat released}}{\text{moles of substance burned}}\).
In our example exercise, burning 5.48 g of formic acid released 30.3 kJ of energy.
First, convert the mass of formic acid to moles using the molar mass (46 g/mol):
  • \(\text{moles of formic acid} = \frac{5.48 \text{ g}}{46 \text{ g/mol}} \approx 0.119 \text{ mol}\).
Subsequently, calculate \(\Delta H\):
  • \(\Delta H = \frac{30.3 \text{ kJ}}{0.119 \text{ mol}} \approx 254.6 \text{ kJ/mol}\).
This calculation tells us how much energy is released per mole of formic acid combusted.Understanding these concepts helps in analyzing exothermic or endothermic reactions.

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Most popular questions from this chapter

Define the heat capacity of a substance. Define the specific heat of a substance.

A 29.1-mL sample of \(1.05 \mathrm{M}\) KOH is mixed with \(20.9 \mathrm{~mL}\) of \(1.07 M \mathrm{HBr}\) in a coffee-cup calorimeter (see Section \(6.6\) of your text for a description of a coffee-cup calorimeter). The enthalpy of the reaction, written with the lowest wholenumber coefficients, is \(-55.8 \mathrm{~kJ} .\) Both solutions are at \(21.8^{\circ} \mathrm{C}\) prior to mixing and reacting. What is the final temperature of the reaction mixture? When solving this problem, assume that no heat is lost from the calorimeter to the surroundings, the density of all solutions is \(1.00 \mathrm{~g} / \mathrm{mL}\), and volumes are additive.

The internal energy of a substance is a state function. What does this mean?

An industrial process for manufacturing sulfuric acid, \(\mathrm{H}_{2} \mathrm{SO}_{4}\), uses hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S}\), from the purification of natural gas. In the first step of this process, the hydrogen sulfide is burned to obtain sulfur dioxide, \(\mathrm{SO}_{2}\). $$ \begin{aligned} 2 \mathrm{H}_{2} \mathrm{~S}(g)+3 \mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{SO}_{2}(g) \\ \Delta H^{\circ} &=-1124 \mathrm{~kJ} \end{aligned} $$ The density of sulfur dioxide at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) is \(2.62 \mathrm{~g} / \mathrm{L}\) and the molar heat capacity is \(30.2 \mathrm{~J} /\left(\mathrm{mol} \cdot{ }^{\circ} \mathrm{C}\right) .\) (a) How much heat would be evolved in producing \(1.00 \mathrm{~L}\) of \(\mathrm{SO}_{2}\) at \(25^{\circ} \mathrm{C}\) and \(1.00\) atm? (b) Suppose heat from this reaction is used to heat \(1.00 \mathrm{~L}\) of \(\mathrm{SO}_{2}\) from \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) to \(500^{\circ} \mathrm{C}\) for its use in the next step of the process. What percentage of the heat evolved is required for this?

A \(5.0-\mathrm{g}\) sample of water starting at \(60.0^{\circ} \mathrm{C}\) loses \(418 \mathrm{~J}\) of energy in the form of heat. What is the final temperature of the water after this heat loss? a. \(20 .{ }^{\circ} \mathrm{C}\) b. \(40 .{ }^{\circ} \mathrm{C}\) c. \(50 .{ }^{\circ} \mathrm{C}\) d. \(60 .{ }^{\circ} \mathrm{C}\) e. \(80 .{ }^{\circ} \mathrm{C}\)
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