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

The enthalpy of formation of \(\mathrm{HCl}(\mathrm{g})\) from the following reaction: \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{HCl}(\mathrm{g})+44 \mathrm{kcal}\) is (a) \(-44\) kcal \(\mathrm{mol}^{-1}\) (b) \(-22\) kcal mol \(^{-1}\) (c) \(22 \mathrm{kcal} \mathrm{mol}^{-1}\) (d) \(-88\) kcal \(\mathrm{mol}^{-1}\)

Short Answer

Expert verified
-22 kcal mol^{-1}

Step by step solution

01

Understand the given reaction and enthalpy change

The given reaction shows the formation of HCl(g) from its elements in their standard states. The reaction also shows that 44 kcal of energy is released in the process, which means the reaction is exothermic.
02

Calculate the enthalpy of formation per mole of HCl

The reaction produces 2 moles of HCl and releases 44 kcal of energy. Therefore, the enthalpy of formation for 1 mole of HCl will be half of 44 kcal.
03

Determine the sign of the enthalpy

Since the reaction is exothermic and energy is released, the enthalpy of formation will have a negative sign, indicating that the system loses energy.

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.

Physical Chemistry
Physical chemistry is a branch of chemistry that deals with how matter behaves on a molecular and atomic level and how chemical reactions occur. It combines principles of physics and chemistry to understand the physical properties of molecules, the forces that act upon them, and the energy changes that occur during chemical reactions. In our context, the study of the enthalpy of formation is a key topic in physical chemistry.

Understanding the enthalpy, or heat content, of substances allows us to predict the energy change that accompanies chemical reactions. It is crucial for the development of new materials, the design of energy-efficient processes, and the understanding of environmental effects caused by chemical reactions. With this knowledge, one can assess the feasibility of industrial chemical processes and optimize them for better performance and lower environmental impact.
Exothermic Reaction
An exothermic reaction is a chemical reaction that releases energy to the surroundings, usually in the form of heat, but also can include light, electricity, or sound. When a reaction is exothermic, the total energy of the products is lower than the total energy of the reactants, thus the change in enthalpy (ablaH) is negative. In the given reaction, the formation of hydrogen chloride gas (HCl(g)) from hydrogen gas (H_{2}(g)) and chlorine gas (Cl_{2}(g)) is exothermic, as indicated by the energy release of 44 kcal. This energy release is a result of the strong new bonds formed in the HCl molecules, which require less energy to maintain compared to the energy stored in the separate H_{2} and Cl_{2} molecules. Therefore, this type of reaction tends to occur spontaneously and is often associated with a rise in temperature of the reaction mixture.
Standard Enthalpy Change
The standard enthalpy change of a reaction, often denoted as Delta H^, refers to the amount of energy absorbed or released during a chemical reaction under standard conditions (typically 25°C and 1 atmosphere of pressure). It is a critical concept for understanding thermochemistry, as it represents the energy change associated with a chemical reaction per mole of reactant or product.

Delta H^ is positive for endothermic reactions (where energy is absorbed from the surroundings) and negative for exothermic reactions (where energy is released). Understanding and calculating the standard enthalpy change allow chemists and chemical engineers to predict whether reactions will favor the formation of products or reactants under standard conditions. In the case of the formation of HCl(g) from H_{2}(g) and Cl_{2}(g), the release of 44 kcal for the production of 2 moles implies a standard enthalpy change of -22 kcal/mol. This value is indicative of an exothermic reaction, revealing that the reaction is energetically favorable and will release energy into the environment, typically making the process self-sustaining once initiated.

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

The factor of \(\Delta G\) values is important in metallurgy. The \(\Delta G\) values for the following reactions at \(800^{\circ} \mathrm{C}\) are given as: \(\mathrm{S}_{2}(\mathrm{~s})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{SO}_{2}(\mathrm{~g}) ; \Delta G=-544 \mathrm{~kJ}\) \(2 \mathrm{Zn}(\mathrm{s})+\mathrm{S}_{2}(\mathrm{~s}) \rightarrow 2 \mathrm{ZnS}(\mathrm{s}) ; \Delta G=-293 \mathrm{~kJ}\) \(2 \mathrm{Zn}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{ZnO}(\mathrm{s}) ; \Delta G=-480 \mathrm{~kJ}\) The \(\Delta G\) for the reaction: \(2 \mathrm{ZnS}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{~g})\) \(\rightarrow 2 \mathrm{ZnO}(\mathrm{s})+2 \mathrm{SO}_{2}(\mathrm{~g})\) will be (a) \(-357 \mathrm{~kJ}\) (b) \(-731 \mathrm{~kJ}\) (c) \(-773 \mathrm{~kJ}\) (d) \(-229 \mathrm{~kJ}\)

The \(\Delta G^{\circ}\) values for the hydrolysis of creatine phosphate (creatine-P) and glucose-6-phosphate \((\mathrm{G}-6-\mathrm{P})\) are (i) Creatine-P \(+\mathrm{H}_{2} \mathrm{O} \rightarrow\) Creatine \(+\mathrm{P}\) \(\Delta G^{\mathrm{o}}=-29.2 \mathrm{~kJ}\) (ii) \(\mathrm{G}-6-\mathrm{P}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{G}+\mathrm{P} ; \Delta G^{\mathrm{o}}=-12.4 \mathrm{~kJ}\) \(\Delta G^{0}\) for the reaction: \(\mathrm{G}-6-\mathrm{P}+\) Creatine \(\rightarrow \mathrm{G}+\) Creatine- \(\mathrm{P}\), is (a) \(+16.8 \mathrm{~kJ}\) (b) \(-16.8 \mathrm{~kJ}\) (c) \(-41.6 \mathrm{~kJ}\) (d) \(+41.6 \mathrm{~kJ}\)

Tungsten carbide is very hard and is used to make cutting tools and rock drills. What is the enthalpy of formation (in \(\mathrm{kJ} / \mathrm{mol}\) ) of tungsten carbide? The enthalpy change for this reaction is difficult of measure directly, because the reaction occurs at \(1400^{\circ} \mathrm{C}\). However, the enthalpies of combustion of the elements and of tungsten carbide can be measured easily. \(2 \mathrm{~W}(\mathrm{~s})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{WO}_{3}(\mathrm{~s}) ; \Delta H\) \(=-1680.6 \mathrm{~kJ}\) \(\mathrm{C}(\) graphite \()+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}) ; \quad \Delta H\) \(=-393.5 \mathrm{~kJ}\) \(2 \mathrm{WC}(\mathrm{s})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{WO}_{3}(\mathrm{~s})+2 \mathrm{CO}_{2}(\mathrm{~g})\) \(\Delta H=-2391.6 \mathrm{~kJ}\) (a) \(-38.0\) (b) \(-76.0\) (c) \(-19.0\) (d) \(-1233.8\)

Enthalpies of solution of \(\mathrm{BaCl}_{2}(\mathrm{~s})\) and \(\mathrm{BaCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{s})\) are \(-20.6 \mathrm{~kJ} / \mathrm{mol}\) and \(8.8 \mathrm{~kJ} / \mathrm{mol}\), respectively. \(\Delta H\) hydration of \(\mathrm{BaCl}_{2}(\mathrm{~s})\) to \(\mathrm{BaCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{s})\) is (a) \(-29.4 \mathrm{~kJ}\) (b) \(-11.8 \mathrm{~kJ}\) (c) \(29.6 \mathrm{~kJ}\) (d) \(11.8 \mathrm{~kJ}\)

Find the bond energy of \(\mathrm{S}-\mathrm{S}\) bond from the following data: \(\mathrm{C}_{2} \mathrm{H}_{5}-\mathrm{S}-\mathrm{C}_{2} \mathrm{H}_{5}(\mathrm{~g}) ; \Delta H_{\mathrm{f}}^{\mathrm{o}}=-148 \mathrm{~kJ}\), \(\mathrm{C}_{2} \mathrm{H}_{5}-\mathrm{S}-\mathrm{S}-\mathrm{C}_{2} \mathrm{H}_{5}(\mathrm{~g}) ; \Delta H_{\mathrm{f}}^{\mathrm{o}}=-202 \mathrm{~kJ}\) \(\mathrm{S}(\mathrm{g}) ; \Delta H_{\mathrm{f}}^{\mathrm{o}}=222 \mathrm{~kJ}\) (a) \(276 \mathrm{~kJ} / \mathrm{mol}\) (b) \(128 \mathrm{~kJ} / \mathrm{mol}\) (c) \(168 \mathrm{~kJ} / \mathrm{mol}\) (d) \(222 \mathrm{~kJ} / \mathrm{mol}\)

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