/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 A sample of \(\mathrm{H}_{2} \ma... [FREE SOLUTION] | 91Ó°ÊÓ

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A sample of \(\mathrm{H}_{2} \mathrm{S}\) gas is placed in an evacuated, sealed container and heated until the following decomposition reaction occurs at \(1000 \mathrm{K} :\) $$2 \mathrm{H}_{2} \mathrm{S}(g) \rightarrow 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \qquad K_{\mathrm{c}}=1.0 \times 10^{-6}$$ As the reaction progresses at a constant temperature of 1000 K, how does the value for the Gibbs free energy constant for the reaction change? (A) It stays constant. (B) It increases exponentially. (C) It increases linearly. (D) It decreases exponentially.

Short Answer

Expert verified
The Gibbs free energy constant for the reaction stays constant. Hence, the correct answer is (A).

Step by step solution

01

Identify the formula

Identify the correct equation that links Gibb's free energy, the temperature and the equilibrium constant. The appropriate equation is \(\Delta G^\circ = -RT \ln K\)
02

Examine the temperature

Examine the statement in the question, 'the reaction progresses at a constant temperature of 1000 K'. This means the variable \(T\) in the equation is constant.
03

Examine the equilibrium constant

One must also note the equilibrium constant \(K_c\), defined as \(K_c = 1.0 \times 10^{-6}\). This value does not change as the reaction progresses, since it is a function of the reaction and temperature, both of which are constant.
04

Conclusion

Since both \(T\) and \(K_c\) are constant, \(\Delta G^\circ \) must also remain constant. The variation in Gibbs free energy is a direct result of changes in either \(T\) or \(K_c\), and as they are continual, so too is \(\Delta G^\circ \).

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Key Concepts

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

Gibbs Free Energy
Gibbs free energy, denoted as \( \Delta G \), is a thermodynamic quantity crucial for determining the spontaneity of a reaction. It involves the balance of enthalpy, entropy, and temperature to predict whether a reaction will proceed without external intervention. The formula for Gibbs free energy change under standard conditions is:\[ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \]where:
  • \( \Delta H^\circ \) is the change in enthalpy,
  • \( T \) is the temperature in Kelvin,
  • \( \Delta S^\circ \) is the change in entropy.
To relate Gibbs free energy specifically to reaction conditions, we often use the formula \( \Delta G = \Delta G^\circ + RT \ln Q \), where \( Q \) is the reaction quotient. However, at equilibrium, this simplifies to \( \Delta G^\circ = -RT \ln K \), with \( K \) being the equilibrium constant. This is pertinent to the exercise, as \( \Delta G \), \( T \), and \( K \) remain constant, indicating that the Gibbs free energy does not change during the reaction.
Reaction Kinetics
Reaction kinetics deals with the speeds or rates at which chemical reactions occur. The decomposition of hydrogen sulfide (\( \mathrm{H}_2\mathrm{S} \)) into hydrogen (\( \mathrm{H}_2 \)) and sulfur (\( \mathrm{S}_2 \)) represents a reaction whose kinetics are essential in understanding how quickly equilibrium is reached.
  • Kinetic factors such as temperature, concentration, and presence of catalysts can influence a reaction rate.
  • The rate law expression, which is derived experimentally, details how these factors affect the speed of the reaction.
Knowing that the reaction temperature is constant, it's clear that kinetic parameters will stay uniform, keeping the rate steady. This kinetic stability supports the condition that the reaction's Gibbs free energy will remain unchanged at equilibrium.
Decomposition Reaction
A decomposition reaction involves a single compound breaking down into two or more simpler substances. In the given exercise, \( 2 \mathrm{H}_2\mathrm{S} \rightarrow 2 \mathrm{H}_2 + \mathrm{S}_2 \) represents such a decomposition.
  • Decomposition reactions are typically endothermic, requiring energy input to break the chemical bonds.
  • These reactions are often influenced by factors such as heat, light, or electricity.
At 1000 K, the decomposition of \( \mathrm{H}_2\mathrm{S} \) progresses under thermal conditions, driving the reaction forward until equilibrium is achieved. Monitoring the decomposition helps us understand why equilibrium constants like \( K_c \) do not change once the reaction stabilizes.
Chemical Equilibrium
Chemical equilibrium occurs when the forward and reverse reactions occur at the same rate, resulting in the concentration of reactants and products remaining constant over time. For the decomposition of \( \mathrm{H}_2\mathrm{S} \), when equilibrium is reached, the rate of formation of \( \mathrm{H}_2 \) and \( \mathrm{S}_2 \) equals the rate of their recombination to form \( \mathrm{H}_2\mathrm{S} \).
  • At equilibrium, the reaction quotient \( Q \) equals the equilibrium constant \( K \), indicating a balance in the reaction.
  • This state of balance means that thermodynamic parameters such as Gibbs free energy are steady.
The concept of equilibrium is crucial for predicting the outcome and positions of chemical reactions in closed systems at constant temperature and pressure, as shown by the constancy of \( \Delta G^\circ \) in the given problem.

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Most popular questions from this chapter

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