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Standard molar enthalpy of formation of \(\mathrm{CO}_{2}\) is equal to (a) standard molar enthalpy of combustion of carbon (graphite) (b) standard molar enthalpy of combustion of gaseous carbon (c) sum of standard molar enthalpies of formation of \(\mathrm{CO}\) and \(\mathrm{O}_{2}\) (d) zero

Short Answer

Expert verified
(a) is correct; CO2 formation from graphite matches the definition.

Step by step solution

01

Understanding Standard Molar Enthalpy of Formation

The standard molar enthalpy of formation is the heat change that results when one mole of a compound is formed from its elements in their standard states under standard conditions (298 K, 1 atm). For CO2, it is the enthalpy change when one mole of CO2 is formed from carbon (graphite) and oxygen gas.
02

Identifying the Reactions for Options

For option (a), the reaction is C (graphite) + O2 (g) → CO2 (g) which is the same reaction defining the standard molar enthalpy of formation of CO2. For option (b), there is no defined standard state for gaseous carbon. Option (c) involves the formation of CO from carbon and its subsequent reaction with O2, which is not standard enthalpy of formation for CO2. Option (d) is incorrect as the enthalpy of formation for a compound from elements in their standard states is not zero unless it is an element itself.
03

Comparing Options with Standard Definition

The standard molar enthalpy of formation of CO2 is actually equal to the standard molar enthalpy of combustion of carbon (graphite), as both involve the formation of CO2 from its elements in their standard states.

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Key Concepts

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

Standard Conditions
In chemistry, standard conditions are a set of agreed-upon parameters that allow scientists to consistently measure and compare chemical reactions and properties. Standard conditions typically refer to a pressure of 1 atm (atmosphere) and a temperature of 298 K (approximately 25°C). These conditions are crucial because they provide a baseline, ensuring that experiments are conducted and observed under similar circumstances, making data reliable and reproducible.
This consistency is especially important for the calculation of enthalpies. Enthalpy changes, such as the standard molar enthalpy of formation, are measured under these standard conditions. This allows chemists to predict and compare the energy changes involved in chemical processes accurately from one study to the next. Consequently, understanding and sticking to standard conditions is essential for anyone venturing into the realm of thermochemistry.
Standard Enthalpy of Combustion
The standard enthalpy of combustion refers to the energy released when one mole of a substance combusts completely with oxygen under standard conditions.
For carbon (graphite), this process can be represented by the chemical equation: C (graphite) + O2 (g) → CO2 (g).
During this reaction, carbon in its most stable form, which is graphite under standard conditions, reacts with oxygen to form carbon dioxide. This reaction releases a specific amount of energy, known as the standard enthalpy of combustion.
  • It is essential to note that the conditions under which this reaction occurs are crucial to correctly calculating and comparing enthalpies.
  • The standard enthalpy of combustion of carbon (graphite) is equivalent to the standard molar enthalpy of formation of CO2 because both present the formation of CO2 from carbon and oxygen under standard conditions.
  • This equivalence is an important concept that helps students understand how these fundamental thermochemical definitions intertwine in certain chemical reactions.
Carbon (Graphite)
Carbon is an element that can exist in various allotropic forms. Graphite is one such form wherein carbon atoms are bonded together in layers of hexagonal lattices.
Graphite is the most stable physical form of carbon under standard conditions. This stability is why graphite is used as the standard state of carbon for thermodynamic studies. It ensures that when we speak about reactions or transformations involving carbon, we use graphite to maintain consistency.
Using graphite as the standard state of carbon is particularly important in defining reactions like the combustion of carbon or the formation of carbon dioxide.
  • In these reactions, graphite serves as the baseline form of carbon from which all measurements originate, ensuring uniformity in data collection and interpretation.
  • It becomes easier to visualize and calculate energy changes when everyone uses the same base reference; in this case, carbon (graphite).
As a result, knowledge about carbon's form—as graphite—and how it behaves under standard conditions is fundamental for comprehending various chemical reactions and their energetic implications.

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Most popular questions from this chapter

Internal energy (E) and pressure of a gas of unit volume are related as (a) \(\mathrm{P}=\frac{2}{3} \mathrm{E}\) (b) \(\mathrm{P}=\frac{3}{2} \mathrm{E}\) (c) \(\mathrm{P}=\frac{\mathrm{E}}{2}\) (d) \(\mathrm{P}=2 \mathrm{E}\)

The standard enthalpy of formation of \(\mathrm{NH}_{3}\) is \(-46.0\) \(\mathrm{kJ} \mathrm{mol}^{-1}\). If the enthalpy of formation of \(\mathrm{H}_{2}\) from its atoms is \(-436 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and that of \(\mathrm{N}_{2}\) is \(-712 \mathrm{~kJ}\) \(\mathrm{mol}^{-1}\), the average bond enthalpy of \(\mathrm{N}-\mathrm{H}\) bond in \(\mathrm{NH}_{3}\) is (a) \(-964 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(+352 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(+1056 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(-1102 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

In monoatomic gases, ratio of specific heat at constant pressure to that at constant volume is (a) \(3 / 5\) (b) \(5 / 3\) (c) \(7 / 5\) (d) \(4 / 5\)

If at \(298 \mathrm{~K}\) the bond energies of \(\mathrm{C}-\mathrm{H}, \mathrm{C}-\mathrm{C}, \mathrm{C}=\mathrm{C}\) and \(\mathrm{H}-\mathrm{H}\) bonds are respectively \(414,347,615\) and \(435 \mathrm{~kJ} \mathrm{~mol}^{-1}\), the value of enthalpy change for the reaction \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{3}(\mathrm{~g})\) at \(298 \mathrm{~K}\) will be (a) \(+250 \mathrm{~kJ}\) (b) \(-250 \mathrm{~kJ}\) (c) \(+125 \mathrm{~kJ}\) (d) \(-125 \mathrm{~kJ}\)

Anhydrous \(\mathrm{AlCl}_{3}\) is covalent. From the data given below, predict whether it would remain covalent or become ionic in aqueous solution (ionization energy of \(\mathrm{Al}=5137 \mathrm{kJmol}^{-1} \Delta \mathrm{H}_{\text {hyuntion }}\) for \(\mathrm{Al}^{+3}=-4665 \mathrm{~kJ}\) \(\mathrm{mol}^{-1}, \Delta \mathrm{H}_{\text {hydation }}\) for \(\left.\mathrm{Cl}^{-}=-381 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)\) (a) ionic (b) covalent (c) both (a) and (b) (d) none of these

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