Question

The enthalpy of formation of ammonia gas is \(-46.0 \mathrm{~kJ} / \mathrm{mol}\). The enthalpy change for the reaction: \(2 \mathrm{NH}_{3}(\mathrm{~g}) \rightarrow \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})\) is (a) \(46.0 \mathrm{~kJ}\) (b) \(92.0 \mathrm{~kJ}\) (c) \(23.0 \mathrm{~kJ}\) (d) \(-92.0 \mathrm{~kJ}\)

Short answer

-92.0 kJ

Step-by-step solution

Determine the Enthalpy Change of Reaction

The enthalpy change for the reaction can be calculated using the enthalpy of formation of ammonia. Since the reaction is the reverse of the formation of ammonia from nitrogen and hydrogen, the enthalpy change for the reaction will be the negative of the enthalpy of formation of ammonia.

Calculate the Enthalpy Change for the Number of Moles Given

Given that the enthalpy of formation of ammonia is -46.0 kJ/mol, reversing the reaction would absorb the same amount of heat. Since there are 2 moles of ammonia reacting, the total enthalpy change will be 2 times -46.0 kJ/mol.

Solve for the Total Enthalpy Change

Multiply the enthalpy change per mole by the number of moles of NH\(_3\) to get the total enthalpy change: \(2 \text{mol} \times -46.0 \text{kJ/mol} = -92.0 \text{kJ}\). Therefore, the correct answer is (d) -92.0 kJ.

Key concepts

Enthalpy of Formation

Understanding the enthalpy of formation is crucial for any student delving into the study of thermochemistry. It refers to the change in enthalpy when one mole of a compound is formed from its elements in their standard states. It's often denoted as \( \Delta H_f^\circ \) and is measured in kilojoules per mole (kJ/mol).

To put it simply, if you think of chemistry as cooking, the enthalpy of formation would be the heat you either need to add or is released when you whip up a specific dish (compound) from its basic ingredients (elements). This concept is pivotal when calculating heat changes in chemical reactions, understanding bond energy, and even in predicting whether reactions will be exothermic or endothermic. For competitive exams, mastering this concept allows you to apply it to various problems, ensuring that calculations of enthalpy changes in reactions are accurate.

Thermochemistry Problems

When dealing with thermochemistry problems, you are essentially analyzing energy changes during chemical reactions. These problems often involve calculating quantities like enthalpy changes, using values such as the enthalpy of formation.

For example, in the given exercise scenario, the reversal of the ammonia formation process represents a core thermochemistry problem. It's not just about applying a formula; it's about understanding what direction a reaction goes, whether energy is absorbed or released, and how to use the stoichiometry of the reaction to calculate total energy changes. Critical thinking and a solid grasp of thermodynamic principles can turn a daunting problem into a series of logical steps leading to the solution.

Tools often used include Hess's law, calorimetry, and the first law of thermodynamics. Practice with varied problems, often presented in competitive exams, ensures a deepened understanding of these concepts and aids in developing the intuition to approach new problems with confidence.

Physical Chemistry for Competitive Exams

For students preparing for competitive exams, physical chemistry can be a game-changer. This branch of chemistry deals with the concepts of energy and matter at a molecular and atomic level. It involves principles that are mathematical and often abstract, requiring not just memorization, but a deep comprehension.

Crucial topics such as enthalpy, entropy, Gibbs free energy, equilibrium, and kinetics are often heavily tested. Preparing effectively means embracing a variety of problems, from straightforward calculations to complex, multi-step scenarios, just like the enthalpy change calculation in our example.

Strengthening your problem-solving skills through regular practice, understanding the underlying principles rather than rote learning, and honing time-management are all vital strategies. In competitive exams, where every mark can set you apart, a rigorous approach to physical chemistry can significantly elevate your performance.