Question

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

Equation (b) represents the standard heat of formation of \( \mathrm{C}_{2} \mathrm{H}_{4} \).

Step-by-step solution

Understand the Standard Heat of Formation

The standard heat of formation for a compound is the change in enthalpy when one mole of the compound is formed from its elements in their standard states. For hydrocarbons like \( \mathrm{C}_{2} \mathrm{H}_{4} \), this involves carbon in its standard state as graphite and hydrogen as \( \mathrm{H}_{2} (\text{g}) \).

Analyze the Elements' Standard States

The standard state of carbon is graphite, not diamond, and for hydrogen, it is \( \mathrm{H}_{2} (\text{g}) \). This means that only equations using graphite and \( \mathrm{H}_{2} (\text{g}) \) as reactants can be considered for standard heat of formation.

Examine Equation Options

Given equations (a) \(2 \mathrm{C} \) (diamond) \(+2 \mathrm{H}_{2}(\text{g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\text{g})\), (b) \(2 \mathrm{C} \) (graphite) \(+2 \mathrm{H}_{2}(\text{g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\text{g})\), (c) \(2 \mathrm{C} \) (diamond) \(+4 \mathrm{H}(\text{g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\text{g})\), and (d) \(2 \mathrm{C} \) (graphite) \(+4 \mathrm{H}(\text{g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\text{g})\), we need graphite and \( \mathrm{H}_{2}(\text{g}) \) to be used.

Identify the Correct Equation

Since the correct standard states for forming \( \mathrm{C}_{2} \mathrm{H}_{4} \) involve graphite and \( \mathrm{H}_{2} \), and not individual hydrogen atoms, equation (b) matches these requirements: \( 2 \mathrm{C} \) (graphite) \(+2 \mathrm{H}_{2}(\text{g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\text{g}) \).

Key concepts

Enthalpy Change

Enthalpy change, commonly denoted as \( \Delta H \), is a crucial concept in chemistry, representing the heat absorbed or released during a chemical reaction at constant pressure. Understanding this change helps chemists determine whether a reaction is endothermic or exothermic.

An endothermic reaction absorbs heat, making \( \Delta H \) positive, while an exothermic reaction releases heat, resulting in a negative \( \Delta H \). The standard heat of formation specifically refers to the enthalpy change when one mole of a compound forms from its elements in their standard states.

Learning to calculate and interpret enthalpy changes is vital for predicting reaction behavior and feasibility, making it an essential part of thermochemistry studies.

Graphite Standard State

The standard state of an element is its most stable form at 1 atm and a specified temperature, usually 25°C (298 K). For carbon, the most stable and therefore standard state is graphite, not diamond.

Graphite is more stable owing to the arrangement of its carbon atoms in layers that provide structural stability.

When discussing standard heats of formation, it's crucial to use elements in their standard states. Hence, using graphite as the form of carbon is necessary for correct calculations and representation of chemical reactions involving carbon.

Hydrocarbon Formation

Hydrocarbon formation involves the chemical process of creating compounds made entirely of hydrogen and carbon. These compounds can be alkanes, alkenes, alkynes, or aromatic hydrocarbons, depending on the types of bonds and the arrangement of carbon atoms.

To form hydrocarbons like ethylene \( \mathrm{C}_{2} \mathrm{H}_{4} \), carbon and hydrogen must react. Typically, carbon in its standard state as graphite, and hydrogen in its molecular form \( \mathrm{H}_{2} (\mathrm{g}) \), come together under the right conditions to synthesize hydrocarbons.

• Alkanes have only single bonds.
• Alkenes, like ethylene, have at least one double bond.
• Alkynes contain triple bonds.

Understanding these formations provides vital insights into fuel production and organic chemistry reactions.

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction in a closed system, leading to constant concentrations of reactants and products. At equilibrium, the system is stable, but reactions continue to occur in both directions.

Le Chatelier's principle helps predict how equilibria shift when external conditions, like temperature or pressure, change.

In the context of hydrocarbon formation, achieving equilibrium is key for optimizing the yield of desired products under specific conditions.
Equilibrium concepts guide process design in laboratories and industries to ensure efficient and cost-effective operations.