Problem 44 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The molar heats of combustion of \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g}), \mathrm{C}\) (graphite) and \(\mathrm{H}_{2}(\mathrm{~g})\) are \(310.62\) kcal, \(94.05\) kcal and \(68.32\) kcal, respectively. Calculate the standard heat of formation of \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g})\)

Step-by-Step Solution

Answer

The standard heat of formation of \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})\) is 54.2 kcal/mol.

Step 1: Identify the Given Information

We are given the molar heats of combustion for acetylene (\(\mathrm{C}_{2} \mathrm{H}_{2}\)), graphite (\(\mathrm{C}\)) and hydrogen (\(\mathrm{H}_{2}\)) as\(\Delta \mathrm{H}_{c} (\mathrm{C}_{2} \mathrm{H}_{2}) = 310.62\) kcal, \(\Delta \mathrm{H}_{c} (\mathrm{C}) = 94.05\) kcal, and \(\Delta \mathrm{H}_{c} (\mathrm{H}_{2}) = 68.32\) kcal, respectively. Our goal is to find the standard heat of formation of acetylene.

Step 2: Write the Balanced Combustion Reactions

For acetylene: \(\mathrm{C}_{2} \mathrm{H}_{2(g)} + \frac{5}{2} \mathrm{O}_{2(g)} \rightarrow 2 \mathrm{CO}_2(g) + \mathrm{H}_2\mathrm{O}(l)\) \(\rightarrow \Delta \mathrm{H}_{c} = -310.62\) kcalFor graphite: \(\mathrm{C} (s) + \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_2 (g)\) \(\rightarrow \Delta \mathrm{H}_{c} = -94.05\) kcalFor hydrogen: \(\mathrm{H}_2 (g) + \frac{1}{2}\mathrm{O}_{2(g)} \rightarrow \mathrm{H}_2\mathrm{O} (l)\) \(\rightarrow \Delta \mathrm{H}_{c} = -68.32\) kcal.

Step 3: Apply Hess's Law

According to Hess's Law, the enthalpy change for a reaction is the sum of the enthalpy changes for the steps involved in the reaction. To find the formation enthalpy for \(\mathrm{C}_{2} \mathrm{H}_{2(g)}\), we rearrange the reactions:- Multiply the combustion reaction of carbon by 2: \(2\times(-94.05)\) kcal- Multiply the combustion reaction of hydrogen by 1: \(68.32\) kcal- Subtract the combustion reaction of acetylene: \(-310.62\) kcal.

Step 4: Calculate the Standard Heat of Formation

Using Hess's Law formula, rearrange the reactions to solve for standard enthalpy of formation:\[ \Delta \mathrm{H}_{f} (\mathrm{C}_{2} \mathrm{H}_{2}) = 2(\Delta \mathrm{H}_{c} (\mathrm{C})) + \Delta \mathrm{H}_{c} (\mathrm{H}_{2}) - \Delta \mathrm{H}_{c} (\mathrm{C}_{2} \mathrm{H}_{2}) \]Insert the given values:\[ \Delta \mathrm{H}_{f} (\mathrm{C}_{2} \mathrm{H}_{2}) = 2(-94.05) + (-68.32) - (-310.62) \]\[ \Delta \mathrm{H}_{f} (\mathrm{C}_{2} \mathrm{H}_{2}) = -188.1 - 68.32 + 310.62 \]\[ \Delta \mathrm{H}_{f} (\mathrm{C}_{2} \mathrm{H}_{2}) = 54.2 \text{ kcal/mol} \]Thus, the standard heat of formation for \(\mathrm{C}_{2} \mathrm{H}_{2}\) is 54.2 kcal/mol.

Key Concepts

Standard Heat of Formation

The standard heat of formation, often symbolized as \( \Delta H_f^\circ \), is a crucial concept in chemistry. It represents the enthalpy change when one mole of a substance is formed from its elements in their most stable states under standard conditions, usually at 1 atm pressure and 298 K temperature. This measurement is significant because it provides a baseline to calculate the energy changes involved in chemical reactions.

Calculating the standard heat of formation involves working with chemical equations and applying thermodynamic principles. In our exercise, the formation of acetylene (\( \mathrm{C}_{2} \mathrm{H}_{2(g)} \)) from its elemental forms, namely carbon (as graphite) and hydrogen, is considered. The reaction would theoretically produce acetylene from graphite and hydrogen gas. Since the formation from elements is not direct but needs to account for energy changes through combustion reactions, Hess's Law becomes a powerful tool to utilize here.

Hess's Law helps in determining the heat of formation by rearranging known reactions that add up to the target formation equation. The overall enthalpy change calculated gives us the standard heat of formation, in this case, 54.2 kcal/mol for acetylene. Remember, each compound has a different standard heat of formation which is pivotal for understanding its energy requirements under ideal conditions.

Molar Heats of Combustion

The molar heats of combustion, denoted as \( \Delta H_c \), is the energy released as heat when one mole of a substance completely reacts with oxygen under standard conditions. It typically results in the formation of carbon dioxide and water when dealing with hydrocarbons. This concept is integral for understanding the energy dynamics of substances and fuels.

• Acetylene: Molar heat of combustion is \( -310.62 \) kcal/mol.
• Graphite: Molar heat of combustion is \( -94.05 \) kcal/mol.
• Hydrogen: Molar heat of combustion is \( -68.32 \) kcal/mol.

The negative signs indicate the exothermic nature of these reactions, where energy is released as bonds form in the combustion products. In our exercise, these values allow us to verify the steps of combustion and use Hess's Law to deduce the enthalpy change leading to the standard heat of formation.

When calculating the standard heat of formation for a compound, molar heats of combustion are combined through stoichiometric coefficients to construct the targeted chemical equation. Then, by summing or subtracting these heats as per Hess's Law, we arrive at the desired formation energy.

Enthalpy Change

Enthalpy change, often represented by \( \Delta H \), is a fundamental measure of heat change at constant pressure in a thermodynamic system. It reflects the quantity of energy absorbed or released during a reaction and can determine whether a reaction is endothermic (+) or exothermic (-).

For chemical reactions, enthalpy changes are pivotal and can be calculated using standard heats such as those of formation or combustion. The ability to calculate and manipulate \( \Delta H \) helps in understanding and predicting the energy flow in reactions. In this exercise, you observed such application by calculating the standard enthalpy of formation using Hess's Law.

• Using the given molar heats of combustion, enthalpy change helped deduce the formation path of acetylene.
• The enthalpy change for each combustion step was used to construct the overall reaction.
• By combining these changes, the standard formation equation enthalpy change could be deduced.

Thus, understanding enthalpy changes not only aids in theoretical calculations but also in practical applications such as energy budgeting for chemical manufacturing.