Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

\(\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) \longrightarrow \mathrm{HS}(\mathrm{g})+\mathrm{H}(\mathrm{g}), \Delta \mathrm{H}^{\circ}=\mathrm{x}_{1}\), \(\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left[\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\right]=\mathrm{x}_{2}, \Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{s}}[\mathrm{H}(\mathrm{g})]=\mathrm{x}_{3}\) hence, \(\Delta \mathrm{H}_{\mathrm{f}}^{\circ}(\mathrm{HS})\) is: (a) \(x_{1}+x_{2}-x_{3}\) (b) \(\mathrm{x}_{3}-\mathrm{x}_{1}-\mathrm{x}_{2}\) (c) \(\mathrm{x}_{1}-\mathrm{x}_{2}-\mathrm{x}_{3}\) (d) \(x_{3}-x_{1}+x_{2}\)

Short Answer

Expert verified
Option (d): \(x_3 - x_1 + x_2\) is correct.

Step by step solution

01

Understand the reaction and given data

The reaction given is the decomposition of \(\text{H}_2\text{S}(\text{g})\) into \(\text{HS}(\text{g})\) and \(\text{H}(\text{g})\). The enthalpy change \(\Delta H^{\circ}\) for this reaction is given as \(x_1\). Additionally, we are provided with the standard enthalpy of formation for \(\text{H}_2\text{S}(\text{g})\) as \(x_2\) and for \(\text{H}(\text{g})\) as \(x_3\). We need to find \(\Delta H_{\mathrm{f}}^{\circ}(\text{HS})\).
02

Apply Hess's Law for enthalpy calculation

According to Hess's Law, the enthalpy change of a reaction \(\Delta H^{\circ}\) can be calculated by considering the enthalpy of formation of the reactants and products. The equation to use here is: \[ \Delta H^{\circ} = \Delta H_{\mathrm{f}}^{\circ}(\text{HS}) + \Delta H_{\mathrm{f}}^{\circ}[\text{H}(\text{g})] - \Delta H_{\mathrm{f}}^{\circ}[\text{H}_2\text{S}(\text{g})] \]Substitute the given values: \[ x_1 = \Delta H_{\mathrm{f}}^{\circ}(\text{HS}) + x_3 - x_2 \]
03

Solve for \(\Delta H_{\mathrm{f}}^{\circ}(\text{HS})\)

Rearrange the equation from Step 2 to isolate \(\Delta H_{\mathrm{f}}^{\circ}(\text{HS})\):\[ \Delta H_{\mathrm{f}}^{\circ}(\text{HS}) = x_1 - x_3 + x_2 \]
04

Match with the answer choices

Compare the derived expression \(\Delta H_{\mathrm{f}}^{\circ}(\text{HS}) = x_1 - x_3 + x_2\) with the given options. It matches option (d).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Enthalpy Change
The enthalpy change, often denoted by \( \Delta H \), is a crucial concept in thermochemistry. It represents the heat absorbed or released during a chemical reaction at constant pressure. When thinking about reactions, it's important to remember:

\[ \Delta H = H_{\text{products}} - H_{\text{reactants}} \]

This formula signifies how the total enthalpy of the products and reactants determines whether a reaction is endothermic (absorbing heat) or exothermic (releasing heat).

  • If \( \Delta H > 0 \), the reaction is endothermic.
  • If \( \Delta H < 0 \), the reaction is exothermic.


Enthalpy change is pivotal in predicting reaction behavior and understanding energy transformations in chemical processes.
Enthalpy of Formation
The enthalpy of formation \( \Delta H_f^{\circ} \) is a specific type of enthalpy change, focused on forming compounds from their constituent elements. For a compound, it is defined as the energy change when one mole of a compound is formed from its elements in their standard states. This standardized value provides a benchmark for comparing how stable a compound is relative to its elements.

For instance, the enthalpy of formation of \( \text{H}_2\text{S}(\text{g}) \) is given as \( x_2 \), and that of \( \text{H}(\text{g}) \) is \( x_3 \). By using these values, together with the enthalpy change of the overall reaction, we apply Hess’s Law. This approach helps find the unknown enthalpy of formation for another product, here being \( \text{HS}(\text{g}) \). To express the unknown formation enthalpy clearly:

\[ \Delta H_{\text{f}}^{\circ}(\text{HS}) = x_1 - x_3 + x_2 \]This showcases how manipulating known formation values can inform us about less common or unknown compounds.
Thermochemistry
Thermochemistry is the study of the heat energy associated with chemical reactions and physical changes. It intricately involves concepts like enthalpy, heat capacity, and phase changes, aiming to understand energy flow in reactions.

Thermochemistry allows scientists to:
  • Predict the energy requirements or releases of reactions.
  • Understand reaction spontaneity and temperature effects.
  • Assess energy conservation and transformation principles.
Mastering thermochemistry involves frequent practice of applying principles like Hess's Law, which states that the total enthalpy change for a reaction is the same, regardless of the pathway or steps of the reaction. Essentially, Hess's Law clarifies how different routes in a reaction mechanism do not alter the overall energy change, facilitating flexibility in calculating unknown enthalpies based upon known data. By adopting Hess’s principle, students can solve complex enthalpy problems with simple algebraic manipulation, integrating both formation and reaction enthalpies into cohesive solutions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free