From the following heats of combustion, $$\begin{array}{r}\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \\\\\Delta H_{\mathrm{rxn}}^{\circ}=-726.4 \mathrm{~kJ} / \mathrm{mol} \\\\\mathrm{C}(\mathrm{graphite})+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-393.5 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \\\\\Delta H_{\mathrm{rxn}}^{\circ}=-285.8 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ calculate the enthalpy of formation of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) from its elements: $$\mathrm{C}(\text { graphite })+2 \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow\mathrm{CH}_{3} \mathrm{OH}(l)$$

Step by step solution

01

Write down the given equations

The heat of combustion for the three following reactions are given: \[\begin{align*}\text{(1)} & \quad \text{CH}_3\text{OH}(l) + \frac{3}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l), \quad \Delta H_{\text{rxn}}^{\circ} = -726.4 \text{ kJ/mol} \\text{(2)} & \quad \text{C}(\text{graphite}) + \text{O}_2(g) \rightarrow \text{CO}_2(g), \quad \Delta H_{\text{rxn}}^{\circ} = -393.5 \text{ kJ/mol} \\text{(3)} & \quad \text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l), \quad \Delta H_{\text{rxn}}^{\circ} = -285.8 \text{ kJ/mol} \end{align*}\]

02

Rearrange the equations to derive the target process

Rearranging equation \(1\) and multiplying equation \(2\) by \(-1\), we get:\[\begin{align*}\text{New (1)} & \quad \text{CH}_3\text{OH}(l) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) + \frac{3}{2}\text{O}_2(g), \quad \Delta H_{\text{rxn}}^{\circ} = 726.4 \text{ kJ/mol} \\text{New (2)} & \quad \text{CO}_2(g) \rightarrow \text{C}(\text{graphite}) + \text{O}_2(g), \quad \Delta H_{\text{rxn}}^{\circ} = 393.5 \text{ kJ/mol} \\text{(3)} & \quad 2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l), \quad \Delta H_{\text{rxn}}^{\circ} = -571.6 \text{ kJ/mol} \end{align*}\]After rearranging, add the three equations to form the desired equation

03

Sum the rearranged equations and enthalpies

Adding these new equations results in the formation equation for methanol, and summing the corresponding enthalpies gives the enthalpy of formation. \[\begin{align*}\text{C}(\text{ graphite }) + 2\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) &\rightarrow \text{CH}_3\text{OH}(l) \\Delta H_{f}^{\circ} &= \Delta H_{\text{New (1)}}^{\circ} + \Delta H_{\text{New (2)}}^{\circ} + \Delta H_{\text{(3)}}^{\circ} \&= 726.4 \text{ kJ/mol} + 393.5 \text{ kJ/mol} - 571.6 \text{ kJ/mol} \&= 548.3 \text{ kJ/mol}\end{align*}\]

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Key Concepts

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

Heats of Combustion

When we speak about the 'heats of combustion,' we're referring to the amount of heat released when a substance burns in the presence of oxygen. This is a crucial concept in thermochemistry because it provides us information about the energy content of fuels and other substances.

To ensure clarity, heats of combustion are typically expressed in kilojoules per mole (kJ/mol) and can be either positive or negative. In most cases, the value is negative, indicating that the reaction is exothermic; this means that energy is released to the surroundings when the substance combusts.

Understanding heats of combustion is fundamental when calculating other thermodynamic properties, such as the enthalpy of formation, through the use of Hess's Law, which brings us to another important topic.

Hess's Law

Hess's Law, named after the Russian chemist Germain Hess, is a principle which states that the total enthalpy change of a chemical reaction is the same no matter how many steps the reaction is carried out in. This is because enthalpy is a state function; it only depends on the initial and final states of the system, not on the pathway taken.

In applying Hess's Law, we can manipulate the equations for known reactions to determine the enthalpy change of a reaction for which the direct measurement is not possible. This law is particularly helpful for calculating the enthalpy of formation of compounds, like in the exercise provided. It involves reversing, multiplying or dividing known equations, and summing their enthalpies accordingly to reach the enthalpy change of the desired reaction.

Thermochemistry

Thermochemistry is the branch of chemistry that deals with the energy and heat associated with chemical reactions and physical transformations. It involves the study of the transfer of energy in the form of heat during chemical changes.

Key concepts in thermochemistry include the first law of thermodynamics (energy cannot be created or destroyed) and exothermic/endothermic processes, which respectively release or absorb heat. For instance, when solving problems related to the enthalpy changes of reactions, understanding exothermic and endothermic reactions is vital for interpreting whether the energy is being absorbed or released.

Chemical Thermodynamics

Chemical thermodynamics is the area of chemistry that studies the interrelation of heat with other forms of energy during chemical processes. It takes into account the principles of thermodynamics to predict the feasibility and extent of chemical reactions.

It involves not just enthalpy, but other state functions such as entropy and free energy, to provide a comprehensive understanding of spontaneous and non-spontaneous reactions. Chemical thermodynamics helps us understand how energy transformations can drive or inhibit chemical reactions, allowing scientists and engineers to predict and control these processes.