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Use Hess's law to determine \(\Delta H^{\circ}\) for the reaction $$\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}), \text { given that }$$ $$\begin{array}{l} \text { C(graphite) }+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g}) \\ &\left.\qquad \Delta H^{\circ}=-110.54 \mathrm{k} \mathrm{J}\right] \end{array}$$ $$\begin{aligned} &\text { C(graphite) }+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})\\\ &&\Delta H^{\circ}=-393.51 \mathrm{kJ} \end{aligned}$$

Short Answer

Expert verified
The standard enthalpy change for the reaction CO(g) + 1/2O2(g) -> CO2(g) is -86.215 kJ.

Step by step solution

01

Analyze the given reactions

First, let's identify the reactions that we are provided with: Reaction 1: C(graphite) +1/2 O2(g) -> CO(g) ΔH1 = -110.54 kJ Reaction 2: C(graphite) + O2(g) -> CO2(g) ΔH2 = -393.51 kJ. The given reactions involve the formation of CO and CO2 from their elements in their standard states, which are the standard enthalpy of formation reactions.
02

Rearrange the reactions

Next, let's manipulate the given reactions to match the target reaction (CO(g) + 1/2O2(g) -> CO2(g)). Notice that the CO2(g) is on the product side in the target reaction and the CO(g) is on the reactant side, opposite to the given reactions. We will make CO(g) from reaction 1 a reactant by reversing the reaction and for reaction 2, we'll keep it as is but divide by 2, to match the 1/2 O2 in the target reaction. So the new reactions are: - Reaction 1: CO(g) -> C(graphite) +1/2 O2(g), ΔH1' = -(ΔH1) = 110.54 kJ - Reaction 2: 1/2 [C(graphite) +O2(g) -> CO2(g)], ΔH2' = 1/2(ΔH2) = -196.755 kJ.
03

Add the reactions

To find the ΔH for the target reaction, simply add the ΔHs from step 2: ΔH = ΔH1' + ΔH2' = 110.54 kJ - 196.755 kJ = -86.215 kJ.

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Key Concepts

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

Enthalpy of Formation
Enthalpy of formation is an essential concept in thermochemistry that helps us understand how energy is involved in creating compounds. It specifically refers to the change in enthalpy when one mole of a compound is formed from its elemental components in their standard states.
To put it simply, imagine building a molecule from scratch using basic atomic "building blocks." The energy change during this process is what's known as the enthalpy of formation.
When examining chemical reactions, like the one involving carbon monoxide (CO) and carbon dioxide (CO₂) from graphite and oxygen, understanding their enthalpies of formation helps determine the overall energy change in the reaction. Enthalpy values are typically measured in kilojoules per mole (kJ/mol), providing a benchmark to foresee if a reaction releases energy (exothermic) or absorbs it (endothermic).
Knowing these values makes it easier to manipulate and calculate Hess's Law thermochemical equations, as they underpin the steps for rearranging and combining reactions to find unknown reaction enthalpies.
Chemical Reactions
Chemical reactions are processes where substances, known as reactants, change into new substances called products. This happens through breaking and forming of chemical bonds, which involves energy changes.
In the context of the given exercise, we're dealing with reactions like
  • Carbon (graphite) and oxygen forming CO
  • Carbon (graphite) and oxygen forming CO₂
Each of these reactions have their own enthalpy changes, as indicated by their corresponding \( \Delta H \) values.
When using Hess's Law, these reactions can be rearranged and combined to find the enthalpy change of a target reaction that might not be easily measurable directly. It ensures that energy, like matter, is conserved, which means the sum of the enthalpies of the individual steps will equal the enthalpy of the overall reaction.
This principle lets chemists construct a clearer picture of energetic pathways in reactions, broadening possibilities for research and practical applications like energy production, material synthesis, and more.
Thermochemistry
Thermochemistry is a branch of chemistry that deals with the study of energy changes, particularly heat, associated with chemical reactions. It provides insight into how reactions absorb or release energy and allows predictions about reaction feasibility and conditions.
Thermochemistry becomes crucial when examining reactions such as the formation of CO and CO₂ from graphite and oxygen. Understanding the thermal energy exchange helps predict and control reactions in industrial, environmental, and biological contexts.
When calculating reaction enthalpies using Hess's Law, thermochemistry provides the framework to calculate energy changes from standard enthalpies of formation. The law's basis is in the first law of thermodynamics, affirming that the energy change in a chemical reaction is constant, regardless of the pathway taken.
This knowledge equips chemists to harness processes effectively, whether that's optimizing fuel efficiency, designing safer chemical plants, or even explaining natural processes like photosynthesis, where such thermochemical principles are at play.

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Most popular questions from this chapter

Given the following information: $$\frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{NH}_{3}(\mathrm{g})\quad\quad\quad\quad\Delta H_{1}^{\circ}$$ $$\mathrm{NH}_{3}(\mathrm{g})+\frac{5}{4} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{NO}(\mathrm{g})+\frac{3}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H_{2}^{\circ}$$ $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\quad\quad\quad\Delta H_{3}^{\circ}$$ Determine \(\Delta H^{\circ}\) for the following reaction, expressed in terms of \(\Delta H_{1}^{\circ}, \Delta H_{2}^{\circ},\) and \(\Delta H_{3}^{\circ}\) $$\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g}) \quad \Delta H^{\circ}=?$$

The combustion of methane gas, the principal constituent of natural gas, is represented by the equation $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+& 2 \mathrm{H}_{2} \mathrm{O}(1) \\ \Delta H^{\circ} &=-890.3 \mathrm{kJ} \end{aligned}$$ (a) What mass of methane, in kilograms, must be burned to liberate \(2.80 \times 10^{7} \mathrm{kJ}\) of heat? (b) What quantity of heat, in kilojoules, is liberated in the complete combustion of \(1.65 \times 10^{4} \mathrm{L}\) of \(\mathrm{CH}_{4}(\mathrm{g})\) measured at \(18.6^{\circ} \mathrm{C}\) and \(768 \mathrm{mmHg} ?\) (c) If the quantity of heat calculated in part (b) could be transferred with \(100 \%\) efficiency to water, what volume of water, in liters, could be heated from 8.8 to \(60.0^{\circ} \mathrm{C}\) as a result?

A calorimeter that measures an exothermic heat of reaction by the quantity of ice that can be melted is called an ice calorimeter. Now consider that \(0.100 \mathrm{L}\) of methane gas, \(\mathrm{CH}_{4}(\mathrm{g}),\) at \(25.0^{\circ} \mathrm{C}\) and \(744 \mathrm{mm} \mathrm{Hg}\) is burned at constant pressure in air. The heat liberated is captured and used to melt \(9.53 \mathrm{g}\) ice at \(0^{\circ} \mathrm{C}\left(\Delta H_{\text {fusion }} \text { of ice }=6.01 \mathrm{kJ} / \mathrm{mol}\right)\) (a) Write an equation for the complete combustion of \(\mathrm{CH}_{4},\) and show that combustion is incomplete in this case. (b) Assume that \(\mathrm{CO}(\mathrm{g})\) is produced in the incomplete combustion of \(\mathrm{CH}_{4}\), and represent the combustion as best you can through a single equation with small whole numbers as coefficients. \((\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) is another . product of the combustion.)

Upon complete combustion, the indicated substances evolve the given quantities of heat. Write a balanced equation for the combustion of \(1.00 \mathrm{mol}\) of each substance, including the enthalpy change, \(\Delta H\), for the reaction.Upon complete combustion, the indicated substances evolve the given quantities of heat. Write a balanced equation for the combustion of \(1.00 \mathrm{mol}\) of each substance, including the enthalpy change, \(\Delta H\), for the reaction. (a) \(0.584 \mathrm{g}\) of propane, \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g}),\) yields \(29.4 \mathrm{kJ}\) (b) \(0.136 \mathrm{g}\) of camphor, \(\mathrm{C}_{10} \mathrm{H}_{16} \mathrm{O}(\mathrm{s}),\) yields \(5.27 \mathrm{kJ}\) (c) \(2.35 \mathrm{mL}\) of acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}(\mathrm{l})(d=0.791\) \(\mathrm{g} / \mathrm{mL}),\) yields \(58.3 \mathrm{kJ}\)

Some of the butane, \(\mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g}),\) in a \(200.0 \mathrm{L}\) cylinder at \(26.0^{\circ} \mathrm{C}\) is withdrawn and burned at a constant pressure in an excess of air. As a result, the pressure of the gas in the cylinder falls from 2.35 atm to 1.10 atm. The liberated heat is used to raise the temperature of 132.5 L of water in a heater from 26.0 to 62.2 ^ C. Assume that the combustion products are \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) exclusively, and determine the efficiency of the water heater. (That is, what percent of the heat of combustion was absorbed by the water?)

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