Calculate the entropy change \((\mathrm{J} / \mathrm{mol} \mathrm{K})\) of the given reaction. The molar entropies [J/K-mol] are given in brackets after each substance. \(2 \mathrm{PbS}(s)[91.2]+3 \mathrm{O}_{2}(g)[205.1] \longrightarrow 2 \mathrm{PbO}(s)[66.5]+2 \mathrm{SO}_{2}(g)[248.2]\) (a) \(-113.5\) (b) \(-168.3\) (c) \(+72.5\) (d) \(-149.2\)

Short Answer

Expert verified
-149.2 J/mol K

Step by step solution

01

Write down the entropies of reactants and products

For the reactants, we have: PbS with entropy of 91.2 J/mol K and O2 with entropy of 205.1 J/mol K. For the products, we have: PbO with entropy of 66.5 J/mol K and SO2 with entropy of 248.2 J/mol K.
02

Calculate the total entropy of reactants

Multiply the given molar entropy of each reactant by its stoichiometric coefficient and then sum these values to find the total entropy of reactants: Total entropy of reactants = (2 mol * 91.2 J/mol K) + (3 mol * 205.1 J/mol K).
03

Calculate the total entropy of products

Multiply the given molar entropy of each product by its stoichiometric coefficient and then sum these values to find the total entropy of products: Total entropy of products = (2 mol * 66.5 J/mol K) + (2 mol * 248.2 J/mol K).
04

Calculate the change in entropy

Subtract the total entropy of reactants from the total entropy of products to find the change in entropy for the reaction: Change in entropy = Total entropy of products - Total entropy of reactants. Ensure the units are consistent, in this case, J/mol K.
05

Solve for the change in entropy

Using the calculations from Steps 2 and 3, solve for the change in entropy: (2 * 66.5 + 2 * 248.2) - (2 * 91.2 + 3 * 205.1) J/mol K.

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Key Concepts

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

Chemical Thermodynamics
Chemical thermodynamics is a sub-discipline of thermodynamics that focuses on the energy changes associated with chemical reactions and changes in physical state. The fundamental principles of thermodynamics such as the first and the second law are applied to chemical processes to predict the spontaneity of reactions, to understand equilibrium conditions, and to calculate various state functions like enthalpy, entropy, and Gibbs free energy.

Understanding the thermodynamics of chemical reactions is crucial because it enables chemists and engineers to design processes that optimize energy use and minimize environmental impact, leading to more sustainable practices. Moreover, concepts like entropy change offer insight into the disorder and randomness within a system, impacting the feasibility and direction of a chemical reaction.
Molar Entropy
Molar entropy is a thermodynamic property representing the amount of disorder or randomness in a substance per mole. Measured in units of joules per mole per kelvin (\( \text{J/mol} \text{K} \)), it's a state function, meaning its value depends only on the state of the system and not how it arrived at that state.

In the context of a reaction, you can consider the molar entropies of the reactants and products to determine if a reaction will result in a higher or lower level of disorder in the system. Generally, an increase in the number of gas molecules or a change from a more ordered solid state to a less ordered gaseous state increases entropy. Conversely, the formation of solids from gases or the decrease in the number of gas molecules reduces entropy.
Physical Chemistry Problems
Problems in physical chemistry often involve calculating the changes in thermodynamic quantities for chemical reactions or phase changes. These include exercises like entropy and enthalpy calculations, determining equilibrium constants, or predicting reaction spontaneity using Gibbs free energy.

For successful problem-solving, it is essential to have a firm grasp of thermodynamic principles and mathematical skills to manipulate equations effectively. Applying these concepts to solve physical chemistry problems not only helps students to understand the fundamental science behind chemical reactions but also hones their analytical skills, which are applicable in various scientific fields.
Stoichiometric Coefficients
Stoichiometric coefficients refer to the numbers written in front of chemical species in a balanced chemical equation. These numbers denote the ratio in which reactants combine and products form. Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction.

In thermodynamic calculations, stoichiometric coefficients are used to determine the quantity of each substance involved in a reaction. They play an essential role in calculating properties like the total entropy of reactants and products, as they tell us how many moles of each substance participate in the reaction. Correctly applying stoichiometric coefficients ensures accurate computation of entropy changes, which is a vital step to predict the direction and spontaneity of a reaction.

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Most popular questions from this chapter

Consider the following reaction at temperature \(T\) : $$ \begin{aligned} &\mathrm{CH}_{2}=\mathrm{CH}_{2}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{ClCH}_{2} \mathrm{CH}_{2} \mathrm{Cl}(g) \\ &\Delta_{r} H^{\circ}=-217.5 \mathrm{~kJ} / \mathrm{mol}, \quad \Delta_{r} S^{\circ}=-233.9 \mathrm{~J} / \mathrm{K}-\mathrm{mol} \end{aligned} $$ Reaction is supported by : (a) entropy (b) enthalpy (c) both (a) \& (b) (d) neither

Using the listed \(\left[\Delta G^{\circ}{ }_{f}\right.\) values \(]\) calculate \(\Delta G^{\circ}\) for the reaction : \(3 \mathrm{H}_{2} \mathrm{~S}(g)[-33.6]+2 \mathrm{HNO}_{3}(l)[-80.6] \stackrel 2 \mathrm{NO}(g)[+86.6]+4 \mathrm{H}_{2} \mathrm{O}(l)[-237.1]+3 S(s)[0.0]\) (a) \(-513.0\) (b) \(-1037.0\) (c) \(+433.4\) (d) \(+225.0\)

One mole of a non-ideal gas undergoes a change of state from \((1.0 \mathrm{~atm}, 3.0 \mathrm{~L}, 200 \mathrm{~K})\) to \((4.0\) atm, \(5.0 \mathrm{~L}, 250 \mathrm{~K}\) ) with a change in internal energy \((\Delta U)=40 \mathrm{~L}\) -atm. The change in enthalpy of the process in L-atm : (a) 43 (b) 57 (c) 42 (d) None of these

Given : \(\mathrm{C}(\) diamond \()+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2} ; \Delta H=-395 \mathrm{~kJ}\) \(\mathrm{C}\) (graphite) \(+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2} ; \Delta H=-393 \mathrm{~kJ}\) The enthalpy of formation of diamond from graphite is (a) \(+2.0 \mathrm{~kJ}\) (b) \(-1.5 \mathrm{~kJ}\) (c) \(-788 \mathrm{~kJ}\) (d) \(788 \mathrm{~kJ}\)

Which of the following expressions is known as Clausius inequality? (a) \(\oint \frac{d q}{T} \leq 0\) (b) \(\oint \frac{d s}{T}=0\) (c) \(\oint \frac{T}{d q} \leq 0\) (d) \(\oint \frac{d q}{T} \geq 0\)

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