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The enthalpy change of a reaction does not depend on (a) initial and final enthalpy change of reaction (b) state of reactants and products (c) different intermediate reactions (d) nature of reactants and products

Short Answer

Expert verified
The enthalpy change does not depend on (c) different intermediate reactions.

Step by step solution

01

Understanding the Concept

The enthalpy change (94H) for a chemical reaction is the difference between the total enthalpy of the products and the total enthalpy of the reactants. According to Hess's Law, the total enthalpy change of a reaction is independent of the pathway taken. This gives us a clue as to what factors really affect the enthalpy change and what factors don't.
02

Analyzing the Options

Let's examine each option provided: (a) Initial and final enthalpy change of reaction - This directly relates to the definition of enthalpy change as it considers the enthalpy difference between reactants and products. (b) State of reactants and products - Since enthalpy is a state function, enthalpy changes depend on whether the substances are in a gaseous, liquid, or solid state at the given conditions. (c) Different intermediate reactions - According to Hess's Law, the overall enthalpy change is independent of the intermediate steps, thus this should not affect the enthalpy change. (d) Nature of reactants and products - The inherent nature of the substances (e.g., molecular structure, bonds) affects their enthalpy.
03

Identifying the Correct Option

Based on the analysis, option (c) 'different intermediate reactions' is the correct answer. This is because, as per Hess's Law, the enthalpy change for a given reaction is the same regardless of the path taken (i.e., the intermediate steps or reactions). Thus, enthalpy change does not depend on these intermediates.

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Key Concepts

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

Hess's Law
Hess's Law is a fundamental principle in thermodynamics that states the overall enthalpy change of a chemical reaction is constant, regardless of the pathway taken from reactants to products. This means a reaction can proceed through multiple steps or just one, and the total enthalpy change will always be the same. This is because enthalpy is a state function. Hess's Law is an invaluable tool in chemistry, particularly when dealing with complex reactions where direct measurement of enthalpy change is difficult.
Using Hess's Law involves several practical applications:
  • Predicting the enthalpy changes of unmeasurable reactions by using known enthalpy changes of related reactions.
  • Validating experimental data related to enthalpy by comparing different reaction pathways.
By understanding Hess's Law, you learn that the overall energy change is dictated purely by the initial and final states, independent of the route taken.
State Function
A state function is a property whose value is determined only by the initial and final states of a system, as opposed to the path or process taken to reach that state. Enthalpy (\(H\)) is an excellent example of a state function. Whether a chemical reaction happens in one step or twenty, as long as it starts and ends in the same place, the enthalpy change stays consistent.
State functions have key characteristics:
  • They depend solely on the current state of the system, not transitional aspects.
  • Other examples besides enthalpy include internal energy, entropy, and pressure.
In problems involving energy changes in reactions, recognizing enthalpy as a state function simplifies the process, allowing chemists to focus purely on the start and end states. The independence from the path imparts robustness to thermodynamic predictions.
Nature of Reactants and Products
The nature of reactants and products directly influences the enthalpy change of a chemical reaction. This involves the chemical identity, bond types, and molecular structure of the substances involved. Though Hess's Law states the pathway doesn't matter, the intrinsic characteristics of these substances do affect the amount of energy required or released.
  • Bond strength: Stronger bonds require more energy to break and release more energy when formed.
  • Molecular structure: Isomers can have different enthalpy values due to their distinct atomic arrangements.
  • Phase of matter: The enthalpy changes when substances transition between gas, liquid, and solid states due to differences in intermolecular forces.
The nature of the reactants and products defines these energy dynamics, making this an essential factor in understanding the energy exchanges in chemical reactions.

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Most popular questions from this chapter

The standard entropies of \(\mathrm{H}_{2}(\mathrm{~g}), \mathrm{I}_{2}(\mathrm{~s})\) and \(\mathrm{HI}(\mathrm{g})\) are \(130.6,116.7\) and \(206.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) respectively. The change in standard entropy in the reaction \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~s}) \longrightarrow 2 \mathrm{HI}(\mathrm{g})\) is (a) \(185.6 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (b) \(170.5 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (c) \(169.5 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (d) \(165.9 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\)

An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If \(\mathrm{T}_{\mathrm{i}}\) is the initial temperature and \(\mathrm{T}_{\mathrm{f}}\) is the final temperature, which of the following statements is correct? (a) \(\left(T_{f}\right)_{i m e v}>\left(T_{i}\right)_{r e v}\) (b) \(\mathrm{T}_{\mathrm{f}}>\mathrm{T}_{\mathrm{i}}\) for reversible process but \(\mathrm{T}_{\mathrm{f}}=\mathrm{T}_{\mathrm{i}}\) for irreversible process (c) \(\left(T_{f}\right)_{\text {irrev }}=\left(T_{i}\right)_{\text {rev }}\) (d) \(\mathrm{T}_{\mathrm{f}}=\mathrm{T}_{\mathrm{i}}\) for both reversible and irreversible processes

For which of the following reactions, is \(\Delta \mathrm{H}\) equal to \(\Delta \mathrm{E} ?\) (a) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{HI}(\mathrm{g})\) (b) \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightarrow \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (c) \(2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\) (d) \(\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})\)

If a gas absorbs \(200 \mathrm{~J}\) of heat and expands by 500 \(\mathrm{cm}^{3}\) against a constant pressure of \(2 \times 10^{\mathrm{s}} \mathrm{Nm}^{-2}\), then change in internal energy is (a) \(-200 \mathrm{~J}\) (b) \(-100 \mathrm{~J}\) (c) \(+100 \mathrm{~J}\) (d) \(+300 \mathrm{~J}\)

The enthalpy changes for the following processes are listed below. \(\mathrm{Cl}_{2}(\mathrm{~g})=2 \mathrm{C} 1(\mathrm{~g}) ; 242.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\mathrm{I}_{2}(\mathrm{~g})=21(\mathrm{~g}) ; 151.0 \mathrm{kJmol}^{-1}\) \(\mathrm{ICl}(\mathrm{g})=\mathrm{I}(\mathrm{g})+\mathrm{Cl}(\mathrm{g}) ; 211.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\mathrm{I}_{2}(\mathrm{~s})=\mathrm{I}_{2}(\mathrm{~g}) ; 62.76 \mathrm{~kJ} \mathrm{~mol}^{-1}\) Given that the standard states for iodine and chlorine are \(\mathrm{I}_{2}\) (s) and \(\mathrm{Cl},(\mathrm{g})\), the standard enthalpy of formation for \(\mathrm{ICl}(\mathrm{g})\) is [2006] (a) \(-14.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(-16.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(+16.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(+244.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

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