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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) \(+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})\) (b) \(2 \mathrm{C}\) (graphite) \(+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})\) (c) \(2 \mathrm{C}\) (diamond) \(+4 \mathrm{H}(\mathrm{g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})\) (d) \(2 \mathrm{C}\) (graphite) \(+4 \mathrm{H}(\mathrm{g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})\)

Short Answer

Expert verified
Equation (b) is correct.

Step by step solution

01

Understand Standard Heat of Formation

The standard heat of formation of a compound is the change in enthalpy when one mole of the compound is formed from its elements in their standard states. For organic compounds like \(\mathrm{C}_{2} \mathrm{H}_{4}\), this typically means using carbon in the form of graphite and hydrogen in its molecular form \(\mathrm{H}_2(\mathrm{g})\).
02

Identify Components in Each Equation

Examine each given equation to determine which elements and their forms are being used:- (a) Carbon is in the form of diamond and hydrogen is \(\mathrm{H}_2(\mathrm{g})\).- (b) Carbon is in the form of graphite and hydrogen is \(\mathrm{H}_2(\mathrm{g})\).- (c) Carbon is in the form of diamond and hydrogen is in atomic form \(\mathrm{H}(\mathrm{g})\).- (d) Carbon is in the form of graphite and hydrogen is in atomic form \(\mathrm{H}(\mathrm{g})\).
03

Standard State of Elements

The standard state of carbon used for formation enthalpy calculations is graphite, and hydrogen should be in its diatomic form \(\mathrm{H}_2(\mathrm{g})\), which means option (b) is using the correct forms.
04

Conclusion

Equation (b) correctly represents the standard heat of formation because it uses carbon in the form of graphite and hydrogen in its molecular form \(\mathrm{H}_2(\mathrm{g})\).

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Key Concepts

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

Enthalpy Change
Enthalpy change refers to the heat energy absorbed or released during a chemical reaction at constant pressure. It's an essential concept when studying thermodynamics and chemistry as it helps determine whether a reaction is exothermic (releases heat) or endothermic (absorbs heat). The standard heat of formation is a specific type of enthalpy change. This is measured under standard conditions, typically at 25°C and 1 atm pressure.
When we say standard enthalpy of formation, we refer to the enthalpy change when one mole of a compound forms from its constituent elements, each in their standard states. Standard states are defined as the most stable physical form of an element at 1 atm and 25°C.
  • An exothermic reaction has a negative enthalpy change as it releases heat to the surroundings.
  • An endothermic reaction has a positive enthalpy change as it absorbs heat.
Understanding enthalpy change is crucial for predicting reaction behavior and energy efficiency.
Graphite
Graphite is a naturally occurring form of carbon that is planar in structure and exhibits excellent electrical conductivity. It is considered the most stable form of carbon and is used as the standard state of carbon for determining enthalpy changes in thermodynamic calculations.
Graphite's stability and structure make it preferable over other forms of carbon, such as diamond, for this purpose. In reactions requiring the standard heat of formation, carbon needs to be in its stable form, which is graphite. This choice ensures consistent and accurate measurements of thermodynamic properties. Selecting the right form of carbon impacts the calculated enthalpy change and the interpretation of reaction energy.
Molecular Hydrogen
Molecular hydrogen, represented as \(\mathrm{H}_2(\mathrm{g})\), is the simplest molecular form of hydrogen and plays a fundamental role in forming organic compounds. It is the standard state of hydrogen due to its diatomic nature, meaning it consists of two hydrogen atoms bonded together.
In chemical reactions, particularly those involving hydrocarbons like ethylene, using molecular hydrogen ensures that calculations of enthalpy change are accurate.
  • Molecular hydrogen is involved in many reactions, from fuel combustion to organic synthesis.
  • It is also important for industrial processes, like the "Haber process" for producing ammonia.
Understanding the properties of molecular hydrogen is vital for various areas of chemistry and engineering.
Organic Compounds
Organic compounds are primarily made up of carbon and hydrogen atoms, often with oxygen, nitrogen, and other elements. These compounds form the basis of life on Earth and include molecules such as proteins, fats, and carbohydrates.
The study of organic compounds, or organic chemistry, focuses on the reactions and structures of these carbon-based molecules. Each organic compound, like \(\mathrm{C}_2\mathrm{H}_4\) (ethylene), has unique properties and potential uses.
  • Organic compounds are crucial for everyday products, from fuels to medicines.
  • The structure of these compounds determines their chemical properties and interactions.
Understanding how these compounds form and react is critical for fields such as pharmaceuticals, agriculture, and energy.

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Most popular questions from this chapter

An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If \(\mathrm{T}_{\mathrm{i}}\) is the initial temperature and \(\mathrm{T}_{\mathrm{f}}\) is the final temperature, which of the following statements is correct? (a) \(\left(T_{f}\right)_{i m e v}>\left(T_{i}\right)_{r e v}\) (b) \(\mathrm{T}_{\mathrm{f}}>\mathrm{T}_{\mathrm{i}}\) for reversible process but \(\mathrm{T}_{\mathrm{f}}=\mathrm{T}_{\mathrm{i}}\) for irreversible process (c) \(\left(T_{f}\right)_{\text {irrev }}=\left(T_{i}\right)_{\text {rev }}\) (d) \(\mathrm{T}_{\mathrm{f}}=\mathrm{T}_{\mathrm{i}}\) for both reversible and irreversible processes

Under the same conditions how many \(\mathrm{mL}\) of \(1 \mathrm{MKOH}\) and \(0.5 \mathrm{MH}_{2} \mathrm{SO}_{4}\) solutions, respectively, when mixed to form total volume of \(100 \mathrm{ml}\), produces the maximum rise in temperature? (a) 67,33 (b) 33,67 (c) 40,60 (d) 50,50

The standard molar enthalpies of formation of cyclohexane (1) and benzene (1) at \(25^{\circ} \mathrm{C}\) are \(-156\) and \(+49 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively. The standard enthalpy of hydrogenation of cyclohexene (1) at \(25^{\circ} \mathrm{C}\) is \(-119 \mathrm{~kJ} /\) mol. Find resonance energy of benzene. (a) \(-152 \mathrm{kJmol}^{-1}\) (b) \(-159 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(+152 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(+159 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

If the bond dissociation energies of \(\mathrm{XY}, \mathrm{X}_{2}\) and \(\mathrm{Y}_{2}\) are in the ratio of \(1: 1: 0.5\) and \(\Delta \mathrm{H}_{f}\) for the formation of \(\mathrm{XY}\) is \(-200 \mathrm{~kJ} / \mathrm{mole}\). The bond dissociation energy of \(\mathrm{X}_{2}\) will be ? (a) \(100 \mathrm{~kJ} / \mathrm{mole}\) (b) \(400 \mathrm{~kJ} / \mathrm{mole}\) (c) \(600 \mathrm{~kJ} / \mathrm{mole}\) (d) \(800 \mathrm{~kJ} / \mathrm{mole}\)

For an ideal gas Joule-Thomson coefficient is (a) positive (b) negative (c) zero (d) dependent on molecular weight

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