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The heats of combustion of carbon and carbon monoxide are \(-393.5\) and \(-283.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\), respectively. The heat of formation (in \(\mathrm{kJ}\) ) of carbon monoxide per mole is : [Main 2016] (a) \(-676.5\) (b) \(-110.5\) (c) \(110.5\) (d) \(676.5\)

Short Answer

Expert verified
The heat of formation of CO per mole is \(-110.5 \text{ kJ/mol}\).

Step by step solution

01

Understand the Problem

We need to find the heat of formation of carbon monoxide per mole (i.e., the enthalpy change for the formation of CO from its elements in their standard states). We are given the heats of combustion for carbon and carbon monoxide.
02

Write the Chemical Equations

Write the equations for the combustion processes: \( \text{C (s) + O}_2 \text{(g)} \rightarrow \text{CO}_2 \text{(g)} \) with \( \Delta H = -393.5 \text{ kJ/mol} \), and \( \text{2CO (g) + O}_2 \text{(g)} \rightarrow \text{2CO}_2 \text{(g)} \) with \( \Delta H = -2 \times 283.5 \text{ kJ/mol} = -567 \text{ kJ/mol} \).
03

Use Hess's Law to Set Up the Reaction for CO Formation

Apply Hess's Law by considering the formation reaction: \( \text{C (s) + 1/2 O}_2 \text{(g)} \rightarrow \text{CO (g)} \). We need to manipulate the given combustion reactions to arrive at this equation.
04

Calculate the Heat of Formation Using Hess's Law

Using the reaction manipulations: 1) Referring the equation \( \text{C (s) + O}_2 \text{(g)} \rightarrow \text{CO}_2 \text{(g)} \) and 2) \( \text{CO (g) + 1/2 O}_2 \text{(g)} \rightarrow \text{CO}_2 \text{(g)} \). To obtain the formation of CO, reverse the second equation: \( \text{CO}_2 \text{(g)} \rightarrow \text{CO (g) + 1/2 O}_2 \text{(g)} \), which is \( +283.5 \text{ kJ/mol} \). Add this to the first equation to get the formation of CO: \(-393.5 \text{ kJ/mol} + 283.5 \text{ kJ/mol} = -110.0 \text{ kJ/mol} \).
05

Select the Correct Answer

Compare the calculated heat of formation \(-110.0 \text{ kJ/mol} \) with the given options. The closest option is \(-110.5 \text{ kJ/mol} \). Therefore, the correct answer is (b) \(-110.5 \text{ kJ/mol}\).

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Key Concepts

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

Hess's Law
To understand Hess's Law, think of it as a powerful tool in chemistry that allows us to calculate the enthalpy changes of reactions by using known enthalpy changes of other reactions. This concept is based on the principle that the total enthalpy change for a reaction is independent of the pathway taken. In simpler terms, if a chemical process can be written as the sum of several steps, the enthalpy change for the total process is the sum of the enthalpy changes for each step.

Hess's Law is especially useful when direct measurement of the enthalpy change of a reaction is difficult. In the case of finding the heat of formation for carbon monoxide (CO), we weren't directly given that information. Instead, we used the combustion reactions of carbon (C) and carbon monoxide (CO) to find it.

The step-by-step manipulation of chemical equations, as seen in the exercise, shows exactly how Hess's Law is applied. By reversing and adding equations to form the desired equation, we could calculate the heat of formation for CO.
Combustion Chemistry
Combustion chemistry involves the burning of a substance in the presence of oxygen, resulting in the production of heat and often light. When we talk about the combustion of carbon or carbon monoxide, we mean their reaction with oxygen to form carbon dioxide and possibly other products.

The heat of combustion is the amount of energy released as heat when a substance undergoes complete combustion with oxygen under standard conditions. For example, in the exercise, the heats of combustion for carbon and carbon monoxide are given as -393.5 kJ/mol and -283.5 kJ/mol, respectively. This represents the energy given off when one mole of these substances burns in oxygen to form carbon dioxide.

An important aspect of combustion chemistry is the understanding of energy released or absorbed during these processes. These heats of combustion were crucial data points for applying Hess's Law to ultimately find the heat of formation of carbon monoxide.
Enthalpy Change
Enthalpy change, denoted as \( \Delta H \), is a key concept in thermodynamics that refers to the heat change of a system at constant pressure. It indicates whether a reaction is exothermic (releases heat) or endothermic (absorbs heat).

In the context of this exercise, we needed to calculate the enthalpy change for the formation of carbon monoxide from its elements. The enthalpy changes provided were for the combustion of substances, not formation, which illustrates how one might often need to rearrange and manipulate reactions to find desired data.

The calculated value for the formation, -110.0 kJ/mol, tells us that producing carbon monoxide from solid carbon and dioxygen is an exothermic process, releasing heat. Accurate calculation of enthalpy changes underpins many aspects of chemistry, allowing us to predict energy needs or yields for industrial processes, laboratory reactions, and even biological systems.

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Most popular questions from this chapter

Show that the reaction \(\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})\) at \(300 \mathrm{~K}\), is spontaneous and exothermic, when the standard entropy change is \(-$$0.094 \mathrm{~kJ} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\). The standard Gibbs free energies of formation for \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) are \(-394.4\) and \(-137.2 \mathrm{~kJ} \mathrm{~mol}^{-1}\), respectively.

Among the following, the intensive property is (properties are) (a) molar conductivity (b) electromotive force (c) resistance (d) heat capacity

The value of \(\log _{10} K\) for a reaction \(A \rightleftharpoons B\) is \(\left(\right.\) Given : \(\Delta_{r} H_{298 \mathrm{~K}}^{\circ}=-54.07 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_{r} S_{298 \mathrm{~K}}^{\circ}\) \(=10 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) and \(R=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\); \(2.303 \times 8.314 \times 298=5705)\) (a) 5 (b) 10 (c) 95 (d) 100

In order to get maximum calorific output, a burner should have an optimum fuel to oxygen ratio which corresponds to 3 times as much oxygen as is required theoretically for complete combustion of the fuel. A burner which has been adjusted for methane as fuel (with \(x\) litre/hour of \(\mathrm{CH}_{4}\) and \(6 x\) litre/hour of \(\mathrm{O}_{2}\) ) is to be readjusted for butane, \(\mathrm{C}_{4} \mathrm{H}_{10}\). In order to get the same calorific output, what should be the rate of supply of butane and oxygen ? Assume that losses due to incomplete combustion, \(e t c\), are the same for both the fuels and the gases behave ideally. Heats of combustion : $$ \mathrm{CH}_{4}=809 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{C}_{4} \mathrm{H}_{10}=2878 \mathrm{~kJ} / \mathrm{mol} $$

An ideal gas is expanded from \(\left(P_{1}, V_{1}, T_{1}\right)\) to \(\left(P_{2}, V_{2}, T_{2}\right)\) under different conditions. The correct statement(s) among the following is (are) (a) The work done on the gas is maximum when it is compressed irreversibly from \(\left(P_{2}, V_{2}\right)\) to \(\left(P_{1}, V_{1}\right)\) against constant pressure \(P_{1}\) (b) If the expansion is carried out freely, it is simultaneously both isothermal as well as adiabatic (c) The work done by the gas is less when it is expanded reversibly from \(V_{1}\) to \(V_{2}\) under adiabatic conditions as compared to that when expanded reversibly from \(V_{1}\) to \(V_{2}\) under isothermal conditions (d) The change in internal energy of the gas is (i) zero, if it is expanded reversibly with \(T_{1}=T_{2}\), and (ii) positive, if it is expanded reversibly under adiabatic conditions with \(T_{1} \neq T_{2}\)

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