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 statement "The change of enthalpy of a chemical reaction is same whether the reaction takes place in one or several steps" is (a) Le Chatelier's law (b) van't Hoff's law (c) first law of thermodynamics (d) Hess's law.

Short Answer

Expert verified
The correct answer is (d) Hess's law.

Step by step solution

01

Identify the Concept

Review each of the given options to ascertain which one correctly describes the statement given.
02

Analyze the Laws

Le Chatelier's law pertains to the shift of equilibrium on changing conditions. Van't Hoff's law relates to osmotic pressure and temperature. The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed in an isolated system. Hess's law states that the total enthalpy change during the complete course of a chemical reaction is the same whether the reaction is made in one step or several steps.
03

Select the Correct Law

By analyzing the statement and comparing it with the laws, it is clear that the statement describes Hess's law, which equates the total change in enthalpy to be independent of the pathway taken by the reaction and instead is reliant only on the initial and final states of the system.

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.

Understanding Enthalpy Change
Enthalpy change is a fundamental concept in thermodynamics relating to the energy changes that occur during chemical reactions. It is defined as the heat absorbed or released by a system at constant pressure. In chemistry, we often use it to predict whether a reaction is endothermic or exothermic.

For an endothermic reaction, heat is absorbed from the surroundings, resulting in a positive enthalpy change, while for an exothermic reaction, heat is released, leading to a negative enthalpy change. Knowing the enthalpy change for specific chemical reactions helps chemists control conditions to optimize reactions for industrial and laboratory processes.

Key Points on Enthalpy Change:

  • It is denoted by the symbol \(\Delta H\).
  • It can be calculated through calorimetry, or estimated using bond energies and standard enthalpies of formation.
  • It plays a crucial role in Hess's law as it allows the calculation of enthalpy changes for a reaction in multiple steps, based on the sum of enthalpy changes for individual steps.
Chemical Reactions and Path Independence
Chemical reactions involve the breaking and forming of bonds between atoms, leading to the transformation of reactants into products. The route taken by a reaction—from reactants to products—can often vary. Some reactions occur directly, while others take multiple intermediate steps. However, according to the concept of state functions, which enthalpy is a part of, the enthalpy change of a chemical reaction is independent of the path taken.

This principle is of great practical importance as it implies that the energy change involved in a chemical reaction can be measured in stages, using multiple intermediates, and the sum will be equivalent to the direct transformation. Thus, the energy profile of a complicated reaction can be broken down into simpler steps for ease of understanding and analysis.

Considering Reaction Pathways:

  • Hess's law exploits the path independence of enthalpy change to calculate it for complex reactions.
  • Knowing the intermediates can provide insights into the reaction mechanism.
First Law of Thermodynamics and Chemical Energy
The first law of thermodynamics, also known as the law of energy conservation, is pivotal in understanding chemical processes. It asserts that energy cannot be created or destroyed in an isolated system—only converted from one form to another. Consequently, during a chemical reaction, the energy of the system and surroundings remains constant; only internal energy distribution changes.

The first law of thermodynamics underpins Hess's law and enthalpy calculations. It assures that all energy changes in the universe are accounted for, whether the energy is in the form of heat, work, or internal chemical energy.

Energy Transformation in Reactions:

  • In any chemical reaction, the change in the internal energy of the system, combined with the work done by or on the system, is equal to the heat absorbed or released.
  • This principle is instrumental in calorimetry, where the heat change associated with reactions can be measured to determine enthalpy changes.

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

Bond dissociation energies of \(\mathrm{H}_{2}, \mathrm{Cl}_{2}\) and \(\mathrm{HCl}_{(g)}\) are 104,58 and \(103 \mathrm{kcal} \mathrm{mol}^{-1}\) respectively. Calculate the enthalpy of formation of \(\mathrm{HCl}\) gas. (a) \(-22 \mathrm{kcal}\) (b) \(+22 \mathrm{kcal}\) (c) \(+184 \mathrm{kcal}\) (d) \(-184 \mathrm{kcal}\)

In a reaction \(P+Q \rightarrow R+S\), there is no change in entropy. Enthalpy change for the reaction \((\Delta H)\) is \(12 \mathrm{~kJ} \mathrm{~mol}^{-1}\). Under what conditions, reaction will have negative value of free energy change? (a) If \(\Delta H\) is positive. (b) If \(\Delta H\) is negative. (c) If \(\Delta H\) is \(24 \mathrm{~kJ} \mathrm{~mol}^{-1}\). (d) If temperature of reaction is high.

The work done during the expansion of a gas from \(4 \mathrm{dm}^{3}\) to \(6 \mathrm{dm}^{3}\) against a constant external pressure of \(3 \mathrm{~atm}\) is \((1 \mathrm{~L} \mathrm{~atm}=101.32 \mathrm{~J})\) (a) \(-6]\) (b) \(-608 J\) (c) \(+304 \mathrm{~J}\) (d) \(-304 \mathrm{~J}\)

What will be the enthalpy of combustion of carbon to produce carbon monoxide on the basis of data given below: $$ \begin{aligned} &\mathrm{C}_{(s)}+\mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)}-393.4 \mathrm{~kJ} \\ &\mathrm{CO}_{(g)}+\frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)}-283.0 \mathrm{~kJ} \end{aligned} $$ (a) \(+676.4 \mathrm{~kJ}\) (b) \(-676.4 \mathrm{~kJ}\) (c) \(-110.4 \mathrm{~kJ}\) (d) \(+110.4 \mathrm{~kJ}\)

Match the column I with column II and mark the appropriate choice. $$\begin{array}{|l|l|l|l|} \hline {\text { Column I }} & & {\text { Column II }} \\ \hline \text { (A) } & \text { State function } & \text { (i) } & \text { At constant pressure } \\ \hline \text { (B) } & \Delta H=q & \text { (ii) } & \text { Specific heat } \\\ \hline \text { (C) } & \Delta U=q & \text { (iii) } & \text { Entropy } \\ \hline \text { (D) } & \text { Intensive property } & \text { (iv) } & \text { At constant volume } \\ \hline \end{array}$$

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