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Write a balanced chemical equation for the formation of \(\mathrm{CH}_{3} \mathrm{OH}(\ell)\) from the elements in their standard states. Find the value for \(\Delta_{f} H^{\circ}\) for \(\mathrm{CH}_{3} \mathrm{OH}(\ell)\) in Appendix L.

Short Answer

Expert verified
The balanced equation is \(\text{C(s)} + 2\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{OH(l)}\), and \(\Delta_f H^{\circ}\) is \(-238.7 \text{ kJ/mol}\).

Step by step solution

01

Identify the Standard States

Methanol ( CH_{3}OH( ext{l} )) is formed from its elements: carbon (C), hydrogen ( H_2 ), and oxygen ( O_2 ). In their standard states, carbon is solid graphite (C(s)), hydrogen is a diatomic gas ( H_2(g) ), and oxygen is a diatomic gas ( O_2(g) ).
02

Write the Skeleton Equation

Combine the elements in their standard states to form methanol. The unbalanced skeletal equation is: \[ \text{C(s)} + \text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{OH(l)} \]
03

Balance the Chemical Equation

Balance the elements in the equation. For methanol formation, one carbon, four hydrogens, and one oxygen are needed, so the balanced equation will look like: \[ \text{C(s)} + 2\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{OH(l)} \]
04

Find \(\Delta_f H^{\circ}\)

Look up the standard enthalpy of formation \(\Delta_f H^{\circ}\) for methanol in Appendix L. The value is \(-238.7 \text{ kJ/mol}\).

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Key Concepts

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

Standard Enthalpy of Formation
The standard enthalpy of formation, often denoted as \( \Delta_f H^{\circ} \), is a crucial concept in chemistry. It refers to the heat change that results when one mole of a compound is formed from its elements in their standard states. This value is typically measured under conditions of 1 atm pressure and a specified temperature, usually 298 K (25 °C).

Understanding \( \Delta_f H^{\circ} \) is important for determining the stability of compounds and predicting reaction outcomes. A negative value indicates that the formation of the compound is exothermic, meaning it releases heat. For methanol (\( \text{CH}_3\text{OH} \)), \( \Delta_f H^{\circ} \) is \(-238.7 \text{ kJ/mol} \), showcasing that forming methanol from its elements releases energy. This released energy implies that methanol is a stable compound under standard conditions.

In summary, standard enthalpies of formation are fundamental for calculating the energetics of chemical reactions, enabling chemists to predict whether a reaction will occur spontaneously or require additional energy.
Standard States of Elements
In chemistry, the standard state of an element is its pure form at 1 atm pressure and a predefined temperature, typically 298 K (25 °C). Recognizing the standard states helps in writing balanced chemical equations and calculating various thermodynamic properties.

Common standard states include:
  • Solid graphite for carbon (\( \text{C(s)} \))
  • Diatomic gas for hydrogen (\( \text{H}_2\text{(g)} \))
  • Diatomic gas for oxygen (\( \text{O}_2\text{(g)} \))
These elements, when used in reactions like the formation of methanol, must be started from their standard states to calculate their heat transformation correctly.

Standard states are independent of the physical state of the element under different conditions; instead, they refer to the most stable form the element takes under standard conditions. Understanding these concepts helps chemists write accurate chemical equations and contextualize chemical data with respect to how elements naturally behave under defined conditions.
Skeletal Chemical Equation
The concept of a skeletal chemical equation is fundamental when starting to balance a chemical reaction. The skeletal equation, also known as an unbalanced equation, outlines the reactants and products without specifying their exact proportions.

To draft a skeletal equation, list the base formulas of the reactants and products. For forming methanol it would look like this: \[ \text{C(s)} + \text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{OH(l)} \]. This unbalanced equation shows which elements are involved but not how many of each type we need.

It's crucial to start with a skeletal equation in chemical reactions because it forms the structure we use to balance the equation. Balancing means adjusting coefficients to ensure the same number of each type of atom appears on both sides of the equation, preserving the law of conservation of mass. Starting with a correctly formed skeletal equation simplifies this balancing process and avoids errors in calculations, particularly those involving thermodynamic properties like \( \Delta_f H^{\circ} \).

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Most popular questions from this chapter

Insoluble \(\mathrm{AgCl}(\mathrm{s})\) precipitates when solutions of \(\mathrm{AgNO}_{3}(\mathrm{aq})\) and \(\mathrm{NaCl}(\mathrm{aq})\) are mixed. \(\mathrm{AgNO}_{3}(\mathrm{aq})+\mathrm{NaCl}(\mathrm{aq}) \rightarrow \mathrm{AgCl}(\mathrm{s})+\mathrm{NaNO}_{3}(\mathrm{aq})\) $$ \Delta_{\mathrm{r}} H^{\circ}=? $$ To measure the energy evolved in this reaction, \(250 . \mathrm{mL}\) of \(0.16 \mathrm{M} \mathrm{AgNO}_{3}(\mathrm{aq})\) and \(125 \mathrm{mL}\) of \(0.32 \mathrm{M} \mathrm{NaCl}(\mathrm{aq})\) are mixed in a coffee-cup calorimeter. The temperature of the mixture rises from \(21.15^{\circ} \mathrm{C}\) to \(22.90^{\circ} \mathrm{C} .\) Calculate the enthalpy change for the precipitation of AgCl(s), in kJ/mol. (Assume the density of the solution is \(1.0 \mathrm{g} / \mathrm{mL}\) and its specific heat capacity is \(4.2 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}\) )

You wish to know the enthalpy change for the formation of liquid \(\mathrm{PCl}_{3}\) from the elements. $$ \mathrm{P}_{4}(\mathrm{s})+6 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{PCl}_{3}(\ell) \quad \Delta_{\mathrm{r}} H^{\circ}=? $$ The enthalpy change for the formation of \(\mathrm{PCl}_{5}\) from the elements can be determined experimentally, as can the enthalpy change for the reaction of \(\mathrm{PCl}_{3}(\ell)\) with more chlorine to give \(\mathrm{PCl}_{5}(\mathrm{s}):\) \(\begin{aligned} \mathrm{P}_{4}(\mathrm{s})+10 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{PCl}_{5}(\mathrm{s}) & \\ \Delta_{r} H^{\circ} &=-1774.0 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn} \\\ \mathrm{PCl}_{3}(\ell)+\mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{PCl}_{5}(\mathrm{s}) & \\ \Delta_{\mathrm{r}} H^{\circ} &=-123.8 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn} \end{aligned}\) Use these data to calculate the enthalpy change for the formation of 1.00 mol of \(\mathrm{PCl}_{3}(\ell)\) from phosphorus and chlorine.

The following terms are used extensively in thermodynamics. Define each and give an example. (a) exothermic and endothermic (b) system and surroundings (c) specific heat capacity (d) state function (e) standard state (f) enthalpy change, \(\Delta H\) (g) standard enthalpy of formation

A piece of chromium metal with a mass of \(24.26 \mathrm{g}\) is heated in boiling water to \(98.3^{\circ} \mathrm{C}\) and then dropped into a coffee-cup calorimeter containing \(82.3 \mathrm{g}\) of water at \(23.3^{\circ} \mathrm{C} .\) When thermal equilibrium is reached, the final temperature is \(25.6^{\circ} \mathrm{C} .\) Calculate the specific heat capacity of chromium.

Assume you mix \(100.0 \mathrm{mL}\) of \(0.200 \mathrm{M} \mathrm{CsOH}\) with \(50.0 \mathrm{mL}\) of \(0.400 \mathrm{M} \mathrm{HCl}\) in a coffee-cup calorimeter. The following reaction occurs: \(\mathrm{CsOH}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{CsCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)\) The temperature of both solutions before mixing was \(22.50^{\circ} \mathrm{C},\) and it rises to \(24.28^{\circ} \mathrm{C}\) after the acid-base reaction. What is the enthalpy change for the reaction per mole of CsOH? Assume the densities of the solutions are all \(1.00 \mathrm{g} / \mathrm{mL}\) and the specific heat capacities of the solutions are \(4.2 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}\)

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