Problem 76 - A Complete Resource Book in Chemistry for JEE Main

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) \(+4 \mathrm{H}(\mathrm{g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})\)

Step-by-Step Solution

Answer

The correct answer is (b).

Step 1: Definition of Standard Heat of Formation

The standard heat of formation is defined as the enthalpy change when one mole of a compound is formed from its elements in their standard states at 1 atmosphere pressure and 298 K. The elements should be in their most stable form.

Step 2: Identification of Stable Forms

The most stable form of carbon under standard conditions is graphite, not diamond. Hydrogen is stable as diatomic hydrogen gas, \(\mathrm{H}_2(\mathrm{g})\).

Step 3: Formation Equation Setup

To determine the correct formation equation, we need \(\mathrm{C}_{2}\) in the form of graphite and \(\mathrm{H}_{2}\) in its gaseous form leading to \(\mathrm{C}_{2}\mathrm{H}_{4}(\mathrm{g})\). This corresponds to \(2 \mathrm{C}(\text{graphite}) + 2 \mathrm{H}_{2}(\mathrm{g}) \rightarrow \mathrm{C}_{2}\mathrm{H}_{4}(\mathrm{g})\).

Step 4: Choose the Correct Option

Based on the above setup, option (b) matches: \(2 \mathrm{C}(\text{graphite}) + 2 \mathrm{H}_{2}(\mathrm{g}) \rightarrow \mathrm{C}_{2}\mathrm{H}_{4}(\mathrm{g})\).

Key Concepts

Enthalpy Change

Enthalpy change refers to the heat absorbed or released during a chemical reaction at constant pressure. Specifically, the standard heat of formation is an enthalpy change occurring when one mole of a compound forms from its elements in their most stable forms.
In chemical reactions, the determination of enthalpy change is crucial because it helps predict whether a reaction will release or absorb energy. This prediction aids in assessing the feasibility and energy efficiency of reactions.

• An exothermic reaction occurs when the enthalpy change is negative, indicating the release of heat.
• Conversely, an endothermic reaction has a positive enthalpy change, signifying heat absorption.

Understanding enthalpy change is vital in thermochemistry for calculating energy requirements and yields for chemical processes.

Stable Form of Elements

The stable form of an element refers to its most energetically favorable state under given conditions, usually at standard temperature and pressure (STP). For calculating the standard heat of formation, these stable forms become the baseline for the reaction.
Carbon, for instance, commonly exists as graphite or diamond, but graphite is more stable under standard conditions. This stability is due to the arrangement of carbon atoms in layers, minimizing energy.

• Graphite is used as the stable form for carbon in formation reactions.
• For hydrogen, the stable form is diatomic hydrogen gas, \\( \mathrm{H}_{2} \mathrm{(g)} \).

Knowing these stable forms is necessary for setting up accurate chemical equations that represent standard conditions.

Chemical Equations

Chemical equations depict the transformation of reactants to products. Correctly setting up a chemical equation involves balancing both the types and the number of atoms on each side of the reaction arrow.
This balance ensures conservation of mass and guides chemists to predict reaction behavior and calculate enthalpy changes accurately.

• A chemical equation for the standard heat of formation needs to have the elements in their stable forms.
• The equation should represent the formation of exactly one mole of the compound.

For example, the standard formation heat of ethylene \\( \mathrm{C}_{2} \mathrm{H}_{4} \) would be represented by \\( 2 \mathrm{C} (\text{graphite}) + 2 \mathrm{H}_2(\mathrm{g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{4} (\mathrm{g}) \). This equation respects the stable forms of the reactants and the need for stoichiometric balance.