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The standard entropies of \(\mathrm{H}_{2}(\mathrm{~g}), \mathrm{I}_{2}(\mathrm{~s})\) and HI (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) \(165.9 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (d) \(165.9 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\)

Short Answer

Expert verified
The change in standard entropy is approximately 165.9 JK⁻¹mol⁻¹ (option d).

Step by step solution

01

Write down the formula for entropy change

The entropy change of a reaction, \( \Delta S^\circ \), is calculated using the formula: \( \Delta S^\circ = \sum S^\circ_{products} - \sum S^\circ_{reactants} \).
02

Identify the entropies for each substance

From the problem, the standard entropies are \( S^\circ_{\mathrm{H}_2(g)} = 130.6 \, \mathrm{JK}^{-1}\mathrm{mol}^{-1} \), \( S^\circ_{\mathrm{I}_2(s)} = 116.7 \, \mathrm{JK}^{-1}\mathrm{mol}^{-1} \), and \( S^\circ_{\mathrm{HI}(g)} = 206.3 \, \mathrm{JK}^{-1}\mathrm{mol}^{-1} \).
03

Calculate the total entropy of reactants

The total entropy of the reactants is the sum of the entropies of \( \mathrm{H}_2(g) \) and \( \mathrm{I}_2(s) \): \( 130.6 + 116.7 = 247.3 \, \mathrm{JK}^{-1}\mathrm{mol}^{-1} \).
04

Calculate the total entropy of products

Since the product is \( 2 \mathrm{HI}(g) \), the total entropy of the products is \( 2 \times 206.3 = 412.6 \, \mathrm{JK}^{-1}\mathrm{mol}^{-1} \).
05

Calculate the change in entropy

Substitute the total entropies of the reactants and products into the entropy change formula: \( \Delta S^\circ = 412.6 - 247.3 = 165.3 \, \mathrm{JK}^{-1}\mathrm{mol}^{-1} \).
06

Choose the closest answer

The closest answer to our calculated change in entropy of \( 165.3 \, \mathrm{JK}^{-1}\mathrm{mol}^{-1} \) given in the options is (d) \( 165.9 \, \mathrm{JK}^{-1}\mathrm{mol}^{-1} \).

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Key Concepts

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

Standard Entropy
Standard entropy is a fundamental concept in thermodynamics, representing the absolute entropy of a substance at a standard state. It is denoted by the symbol \( S^\circ \) and is measured in units of \( \text{J K}^{-1} \text{mol}^{-1} \). Each substance has a unique standard entropy value which can be found in tables for substances under standard conditions (usually 298.15 K and 1 atm pressure).
Standard entropy helps us understand the degree of disorder or randomness in a system. For instance, gases generally have higher standard entropy values compared to their solid or liquid counterparts due to greater molecular movement and disorder.
In chemical reactions, the difference between the standard entropies of the products and the reactants gives us insight into how the disorder of the system changes.
Chemical Reaction
A chemical reaction involves the transformation of one or more substances into new substances. It is represented by a balanced chemical equation, which includes the reactants (initial substances) and the products (new substances formed).
In the context of the original exercise, the chemical reaction is \( \mathrm{H}_2(\mathrm{g})+\mathrm{I}_2(\mathrm{s}) \longrightarrow 2 \mathrm{HI}(\mathrm{g}) \). This means hydrogen gas reacts with iodine solid to produce hydrogen iodide gas.
This reaction changes both the chemical bonds and the entropy of the system, which we assess using standard entropies. Understanding the chemical equation is crucial as it guides us in determining the amounts of each reactant and product involved, necessary for entropy calculations.
Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. It is heavily involved in chemical reactions, particularly in analyzing energy changes and entropy shifts.
There are four laws of thermodynamics, but for chemical reactions, the second law is often the most relevant. It states that the total entropy of an isolated system always increases over time, aligning with the idea that systems tend to evolve towards a state of greater disorder.
In the given exercise, we use thermodynamic principles to calculate the change in standard entropy, informing us about the disorder increase in the transformation from reactants to products. Understanding these principles helps explain why reactions proceed and how they relate to energy changes and system balance.
Entropy Calculation
Calculating entropy change in a chemical reaction involves comparing the standard entropies of the products and reactants. The formula used is: \[ \Delta S^\circ = \sum S^\circ_{\text{products}} - \sum S^\circ_{\text{reactants}} \].
To use this formula, follow these steps:
  • Identify each substance's standard entropy from the given data.
  • Multiply the standard entropy of each substance by its coefficient in the balanced equation to find the total entropy for reactants and products.
  • Subtract the total entropy of reactants from the total entropy of products to find the change.
In the exercise, the calculation is applied as follows: the sum of entropies for two moles of \( \mathrm{HI} \) is subtracted from the sum of entropies of \( \mathrm{H}_2 \) and \( \mathrm{I}_2 \). This straightforward method helps us quantify the disorder change in the process, critical for predicting reaction spontaneity.

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Most popular questions from this chapter

The internal energy change when a system goes from state \(\mathrm{A}\) to \(\mathrm{B}\) is \(40 \mathrm{~kJ} / \mathrm{mol}\). If the system goes from \(\mathrm{A}\) to \(B\) by a reversible path and returns to state \(A\) by an irreversible path what would be the net change in internal energy? (a) \(40 \mathrm{~kJ}\) (b) \(>40 \mathrm{~kJ}\) (c) \(<40 \mathrm{~kJ}\) (d) zero

Heat required to raise the temperature of \(1 \mathrm{~mol}\) of a substance by \(1^{\circ}\) is called: (a) Specific heat (b) Molar heat capacity (c) Water equivalent (d) Specific gravity

\(0.16 \mathrm{~g}\) of methane is subjected to combustion at \(27^{\circ} \mathrm{C}\) in a bomb calorimeter system. The temperature of the calorimeter system (including water) was found to rise by \(0.5^{\circ} \mathrm{C}\). Calculate the heat of combustion of methane at constant volume. The thermal capacity of the calorimeter system is \(177 \mathrm{~kJ} \mathrm{~K}^{-1}\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)\) : (a) \(-695 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(-1703 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(-890 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(-885 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Which of the following equations represent standard heat of formation of \(\mathrm{C}_{2} \mathrm{H}_{4} ?\) (a) \(2 \mathrm{C}\) (diamond) \(+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})\) (b) \(2 \mathrm{C}\) (graphite) \(+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})\) (c) \(2 \mathrm{C}\) (diamond) \(+4 \mathrm{H}(\mathrm{g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})\) (d) \(2 \mathrm{C}\) (graphite) \(+4 \mathrm{H}(\mathrm{g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})\)

Standard entropy of \(\mathrm{X}_{2}, \mathrm{Y}_{2}\) and \(\mathrm{XY}_{3}\) are 60,40 and 50 \(\mathrm{JK}^{-1} \mathrm{~mol}^{-1}\), respectively. For the reaction: \(1 / 2 \mathrm{X}_{2}+3 / 2 \mathrm{Y}_{2} \longrightarrow \mathrm{XY}_{3}, \Delta \mathrm{H}=-30 \mathrm{~kJ}\), to be at equilibrium, the temperature will be: (a) \(1250 \mathrm{~K}\) (b) \(500 \mathrm{~K}\) (c) \(750 \mathrm{~K}\) (d) \(1000 \mathrm{~K}\)

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