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Chapter 5: Problem 48

The reaction of \(2 \mathrm{~mol} \mathrm{H}_{2}(\mathrm{~g})\) with \(1 \mathrm{~mol} \mathrm{O}_{2}(\mathrm{~g})\) to yield \(2 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}(\ell)\) is exothermic, with \(\Delta H=-572 \mathrm{~kJ} .\) Calcu- late the enthalpy change observed when \(10.0 \mathrm{~g} \mathrm{O}_{2}(\mathrm{~g})\) reacts with an excess of hydrogen.

Short Answer

Expert verified
-178.75 kJ

Step by step solution

01

Find Moles of Oxygen

First, we need to determine the number of moles of oxygen gas (\(\mathrm{O}_2\)) in the 10.0 g sample. Use the molar mass of \(\mathrm{O}_2\), which is approximately 32.00 g/mol.\[\text{Moles of } \mathrm{O}_2 = \frac{10.0 \text{ g}}{32.00 \text{ g/mol}} = 0.3125 \text{ mol}\]
02

Assess Reaction Based on Stoichiometry

From the balanced chemical equation, \[2 \mathrm{H}_2 (g) + 1 \mathrm{O}_2 (g) \rightarrow 2 \mathrm{H}_2 \mathrm{O} (\ell)\], 1 mole of \(\mathrm{O}_2\) reacts to release \(-572\, \mathrm{kJ}\). We will use this stoichiometry to find the enthalpy change for the given reaction.
03

Calculate Enthalpy Change for Given Reaction

Now, calculate the enthalpy change when 0.3125 moles of \(\mathrm{O}_2\) react. Since 1 mole of \(\mathrm{O}_2\) releases \(-572\, \mathrm{kJ}\), use the following proportion:\[\Delta H = 0.3125 \text{ mol} \times \left(-572\, \mathrm{kJ/mol}\right) = -178.75\, \mathrm{kJ}\]
04

Conclusion

The enthalpy change observed when 10.0 g of \(\mathrm{O}_2\) reacts with an excess of hydrogen is \(-178.75\, \mathrm{kJ}\). This indicates that the process releases this amount of energy as heat.

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Key Concepts

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

Stoichiometry
Stoichiometry is an essential concept in chemistry. It involves the calculation of reactants and products in chemical reactions. It is based on the principle of conservation of mass. This means the total mass of reactants equals the total mass of products.
In the reaction given, stoichiometry allows us to determine how much heat is released when a certain mass of oxygen reacts. The balanced equation \(2 \mathrm{H}_2 (g) + 1 \mathrm{O}_2 (g) \rightarrow 2 \mathrm{H}_2 \mathrm{O} (\ell)\) shows the stoichiometric ratio.
Here鈥檚 how stoichiometry works:
  • The coefficients in a balanced equation indicate the number of moles of each substance involved in the reaction.
  • For every 1 mole of \(\mathrm{O}_2\), 2 moles of \(\mathrm{H}_2\) are required. The reaction produces 2 moles of \(\mathrm{H}_2\mathrm{O}\).
  • This stoichiometric relationship helps calculate the energy change in reactions.
Exothermic Reaction
An exothermic reaction is one that releases energy in the form of heat. This is a common feature in many combustion and oxidation reactions.
Let's look at the reaction \(2 \mathrm{H}_2 (g) + 1 \mathrm{O}_2 (g) \rightarrow 2 \mathrm{H}_2 \mathrm{O} (\ell)\). In this reaction, the conversion of hydrogen and oxygen into water releases energy, characterized by a negative enthalpy change \(\Delta H\).
Important points about exothermic reactions:
  • Exothermic reactions release heat to the surroundings, making them feel hot.
  • The negative sign of \(\Delta H = -572 \mathrm{~kJ/mol}\) for this reaction indicates an exothermic process, showcasing the energy released with one mole of \(\mathrm{O}_2\).
  • Exothermic reactions are energetically favorable, often occurring spontaneously.
Molar Mass Calculation
Molar mass calculation is a fundamental skill in stoichiometry. It allows for the conversion between grams and moles, aiding in the description of amounts of substances involved in chemical reactions.
When calculating molar mass, you sum the atomic masses of each element in a compound, as shown by the periodic table. For oxygen gas \(\mathrm{O}_2\), the molar mass can be calculated as:
  • Oxygen's atomic mass is approximately 16.00 g/mol.
  • For \(\mathrm{O}_2\), multiply by 2, resulting in \(32.00 \mathrm{~g/mol}\).
The formula \(\text{Moles of } \mathrm{O}_2 = \frac{\text{mass}}{\text{molar mass}}\) allows us to find the number of moles from a given mass of substance.
By applying this knowledge, we can accurately determine how many moles are involved in a reaction, thereby calculating the enthalpy change for reaction processes.

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Most popular questions from this chapter

Ammonia is produced commercially by the direct reaction of the elements. The formation of \(5.00 \mathrm{~g}\) gaseous \(\mathrm{NH}_{3}\) by this reaction releases \(13.56 \mathrm{~kJ}\) of heat. $$ \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) $$ (a) What is the sign of the enthalpy change for this reaction? (b) Calculate \(\Delta H\) for the reaction, assuming molar amounts of reactants and products.

You want to heat the air in your house with natural gas \(\left(\mathrm{CH}_{4}\right)\). Assume your house has \(275 \mathrm{~m}^{2}\) (about \(2960 \mathrm{ft}^{2}\) ) of floor area and that the ceilings are \(2.50 \mathrm{~m}\) from the floors. The air in the house has a molar heat capacity of \(29.1 \mathrm{~J} / \mathrm{mol} \mathrm{K}\). (The number of moles of air in the house can be found by assuming that the average molar mass of air is \(28.9 \mathrm{~g} / \mathrm{mol}\) and that the density of air at these temperatures is \(1.22 \mathrm{~g} / \mathrm{L} .)\) What mass of methane do you have to burn to heat the air from \(15.0^{\circ} \mathrm{C}\) to \(22.0^{\circ} \mathrm{C}\) ?

Explain why absolute enthalpies cannot be measured and only changes can be determined.

The law of Dulong and Petit states that the heat capacity of metallic elements is approximately \(25 \mathrm{~J} / \mathrm{mol} \cdot{ }^{\circ} \mathrm{C}\) at \(25^{\circ} \mathrm{C}\). In the 19 th century, scientists used this relationship to obtain approximate atomic masses of metals, from which they determined the formulas of compounds. Once the formula of a compound of the metal with an element of known atomic mass is known, the mass percentage composition of the compound is used to find the atomic mass of the metal. The following example shows the calculations involved. (a) Experimentally, the specific heat of a metal is found to be \(0.24 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\). Use the law of Dulong and Petit to calculate the approximate atomic mass of the metal. (b) An oxide of this element is \(6.90 \%\) oxygen by mass. Use the molar mass of \(16.00 \mathrm{~g} / \mathrm{mol}\) for oxygen and the approximate atomic mass found in part (a) to determine the subscripts \(x\) and \(y\) in the formula of the oxide, \(\mathrm{M}_{x} \mathrm{O}_{y} .\) (The mole ratio of the elements you find will not be exactly whole numbers, so considerable rounding is needed to obtain whole numbers in the formula.) (c) From the formula established in part (b), \(x\) mol \(M\) are combined with \(y\) mol \(O .\) Calculate the mass of the metal that is combined with \(y \mathrm{~mol} \mathrm{O}\), using the percent composition of the oxide, and find the atomic mass of the metal. What is the element \(\mathrm{M}\) ?

A 1: 1 mole ratio of \(\mathrm{CO}(\mathrm{g})\) and \(\mathrm{H}_{2}(\mathrm{~g})\) is called water gas. It is used as a fuel because it can be burned in air: $$ 2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g}) \quad \Delta H=-566 \mathrm{~kJ} $$ \(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\ell) \quad \Delta H=-571.7 \mathrm{~kJ}\) (a) Find the number of moles of \(\mathrm{CO}(\mathrm{g})\) and \(\mathrm{H}_{2}(\mathrm{~g})\) present in \(10.0 \mathrm{~g}\) water gas. (Remember that they are present in a 1: 1 mole ratio.) (b) Use the preceding thermochemical equations to find the enthalpy change when \(10.0 \mathrm{~g}\) water gas is burned in air.
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