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Which of the following equations represent standard heat of formation of \(\mathrm{C}_{2} \mathrm{H}_{4} ?\) a. \(2 \mathrm{C}\) (diamond) \(+2 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})\) b. \(2 \mathrm{C}\) (graphite) \(+2 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})\) c. \(2 \mathrm{C}\) (diamond) \(+4 \mathrm{H}(\mathrm{g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})\) d. \(2 \mathrm{C}\) (graphite) \(+4 \mathrm{H}(\mathrm{g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})\)

Short Answer

Expert verified
Option (b) is correct: \(2 \mathrm{C} \text{ (graphite)} + 2 \mathrm{H}_{2}(\mathrm{g}) \rightarrow \mathrm{C}_{2}\mathrm{H}_{4}(\mathrm{g})\).

Step by step solution

01

Understand the Concept of Heat of Formation

The standard heat of formation for a compound is the enthalpy change when one mole of the compound is formed from its elements in their standard states. For ethylene (\(\mathrm{C}_{2}\mathrm{H}_{4}\)), we want to form it from carbon and hydrogen in their standard forms, which are graphite for carbon and diatomic hydrogen molecules for hydrogen.
02

Identify the Correct Standard States

Carbon has two allotropes, graphite and diamond, but graphite is the standard state. Hydrogen exists as diatomic molecules (\(\mathrm{H}_{2}(\mathrm{g})\)) in its standard state. Therefore, to find the correct formation equation, we must have \(\mathrm{C}(\text{graphite})\) and \(\mathrm{H}_{2}(\mathrm{g})\) as reactants.
03

Equation Analysis

Check each equation: - (a) uses diamond and diatomic hydrogen. - (b) uses graphite and diatomic hydrogen. - (c) uses diamond and atomic hydrogen. - (d) uses graphite and atomic hydrogen. Only equation (b) uses both graphite and diatomic hydrogen, making it consistent with the definition of standard heat of formation.
04

Conclusion

From the analysis, the equation that correctly represents the standard heat of formation of \(\mathrm{C}_{2}\mathrm{H}_{4}\) is option (b), as it uses both carbon in its standard form (graphite) and diatomic hydrogen gas.

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Key Concepts

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

Enthalpy Change
The concept of enthalpy change is core to understanding many chemical reactions, including the standard heat of formation. Enthalpy change, represented by \( \Delta H \), is the total heat content that changes during a reaction, which can either be absorbed or released. It is usually measured in kilojoules per mole (kJ/mol).
The standard heat of formation is a specific type of enthalpy change. It measures the heat change when one mole of a compound is formed from its pure elements under standard conditions. These conditions typically involve a pressure of 1 bar and a temperature of 298 K (25°C).
Knowing the enthalpy changes helps us predict the energy required or released in forming compounds, like ethylene \( \mathrm{C}_{2}\mathrm{H}_{4} \), from their elemental forms.
Standard States
Standard states refer to the most stable physical form of an element or compound at a set standard physical condition, generally accepted at 1 bar and 298 K (25°C). The role of standard states is essential in defining enthalpies like the standard heat of formation. Broad consistency in these states allows accurate heat measurement across reactions.
For instance, when we consider carbon, its standard state at room temperature is graphite, which is the most stable form compared to diamond. Hydrogen, another key example in our context, naturally exists as diatomic molecules \( \mathrm{H}_{2}(\mathrm{g}) \) under standard conditions.
  • Graphite for Carbon
  • Diatomic \( \mathrm{H}_{2}(\mathrm{g}) \) for Hydrogen
Therefore, using correct standard states in chemical equations is crucial for applying the concept of standard heat of formation accurately.
Allotropes of Carbon
Carbon is an extraordinary element that exists in different structural forms known as allotropes, which include graphite, diamond, and several others like fullerenes and graphene. In a chemistry context, understanding these forms is vital because they impact properties like stability and reactivity.
Graphite and diamond are the most commonly discussed allotropes. Graphite is the most stable at room temperature, making it the standard state of carbon for calculations involving standard heat of formation. It's composed of layers of carbon atoms in a hexagonal arrangement, allowing it to easily slide over each other, accounting for properties such as lubricity.
When considering reactions involving carbon, one must choose the right allotrope. For standard heat of formation, graphite is the correct choice.
Diatomic Molecules
Diatomic molecules consist of two atoms that can either be the same or different chemical elements. Diatomic molecules are significant in chemistry because many elements exist in this form under standard conditions.
One prominent example is hydrogen which exists as \( \mathrm{H}_{2} \), a diatomic molecule. This means, under normal conditions, hydrogen pairs up to form two-atom clusters.
  • Common diatomic molecules include: \( \mathrm{H}_{2} \), \( \mathrm{N}_{2} \), \( \mathrm{O}_{2} \), \( \mathrm{F}_{2} \), \( \mathrm{Cl}_{2} \), \( \mathrm{Br}_{2} \), \( \mathrm{I}_{2} \).
Understanding and correctly identifying diatomic molecules is essential, especially while calculating standard heat of formation, since it involves elements in their standard states. The diatomic nature affects both the chemical equations used and the enthalpy values applied.

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Most popular questions from this chapter

Among the following, the state function(s) is/are a. Internal energy b. Irreversible expansion work c. Reversible expansion work d. Molar enthalpy

In a constant volume calorimeter, \(3.5\) of a gas with molecular weight 28 was burnt in excess oxygen at \(298.0 \mathrm{~K}\) The temperature of the calorimeter was found to increase from \(298.0 \mathrm{~K}\) to \(298.45 \mathrm{~K}\) due to the combustion process. Given that the heat capacity of the calorimeter is \(2.5 \mathrm{~kJ} \mathrm{~K}^{-1}\), the numerical value for the enthalpy of combustion of the gas in \(\mathrm{kJ} \mathrm{mol}^{-1}\) is a. 10 b. 100 c. 5 d. 95 [IIT 2009]

Calculate \(\Delta \mathrm{G}^{\circ}\) for the reaction below and tell whether it is spontaneous or non-spontaneous under standard conditions at \(25^{\circ} \mathrm{C}\). \(2 \mathrm{~S}(\mathrm{~s})+3 \mathrm{O}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{H}_{2} \mathrm{SO}_{4}(l)\) \(\Delta \mathrm{H}^{\circ}=-1056 \mathrm{~kJ} / \mathrm{mol}\) \(\Delta \mathrm{S}^{0}=-505 \mathrm{~J} / \mathrm{mol}\) a. \(\Delta \mathrm{G}^{\circ}=-906 \mathrm{~kJ}\) and the process is spontaneous b. \(\Delta \mathrm{G}^{0}=-1206\) and the process is spontaneous c. \(\Delta \mathrm{G}^{\circ}=-906 \mathrm{~kJ}\) and the process is non- spontaneous d. \(\Delta \mathrm{G}^{0}=-1206 \mathrm{~kJ}\) and the process is non- spontaneous

The solubility of manganese (II) fluoride in water is \(6.6 \mathrm{~g} / \mathrm{ml}\) at \(40^{\circ} \mathrm{C}\) and \(4.8 \mathrm{~g}\) /litre at \(100^{\circ} \mathrm{C}\). Based on this data, what is the sign of \(\Delta \mathrm{H}^{\circ}\) and \(\Delta \mathrm{S}^{\circ}\) for the following process? \(\mathrm{MnF}_{2}(\mathrm{~s}) \Rightarrow \mathrm{Mn}^{2+}(\mathrm{aq})+2 \mathrm{~F}^{-}(\mathrm{aq})\) a. \(\Delta \mathrm{H}^{\circ}\) is negative and \(\Delta \mathrm{S}^{\circ}\) is positive b. \(\Delta \mathrm{H}^{\circ}\) is negative and \(\Delta \mathrm{S}^{\circ}\) is negative c. \(\Delta \mathrm{H}^{\circ}\) is positive and \(\Delta \mathrm{S}^{\circ}\) is positive d. \(\Delta \mathrm{H}^{\circ}\) is positive and \(\Delta \mathrm{S}^{\circ}\) is negative

Which of the following is/are spontaneous process? a. \(2 \mathrm{NH}_{3}(\mathrm{~g}) \rightarrow \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})\) if \(\mathrm{p}_{\mathrm{NH}_{3}}=1 \mathrm{~atm}\), \(\mathrm{p}_{\mathrm{H}_{2}}=\mathrm{P}_{\mathrm{N}_{2}}=0\) and \(\mathrm{Kp}=2 \times 10^{-6}\) b. \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) if \(\mathrm{pH}_{2}=\mathrm{PN}_{2}=1\) atm, \(\mathrm{pNH}_{3}=0\) and \(\mathrm{Kp}=4 \times 10^{5}\) c. the expansion of gas into a vacuum d. dissolving more solute in a saturated solution

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