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Indicate whether \(\Delta G\) increases, decreases, or does not change when the partial pressure of \(\mathrm{H}_{2}\) is increased in each of the following reactions: (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) \(2 \mathrm{HBr}(\mathrm{g}) \longrightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{Br}_{2}(g)\) (c) \(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})\)

Short Answer

Expert verified
(a) \(\Delta G\) increases, (b) \(\Delta G\) decreases, (c) \(\Delta G\) increases as the partial pressure of \(\mathrm{H}_{2}\) is increased in each reaction.

Step by step solution

01

Reaction (a):

\(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\). For this reaction, the reaction quotient \(Q\) can be expressed as: \[Q = \frac{[\mathrm{NH}_{3}]^{2}}{[\mathrm{N}_{2}][\mathrm{H}_{2}]^{3}}\] Now, we need to see the effect of increasing the partial pressure of \(\mathrm{H}_{2}\) on \(Q\). Since the partial pressure of \(\mathrm{H}_{2}\) is in the denominator of the expression for \(Q\), if its value increases, the value of \(Q\) must decrease. This implies that: \( \Delta G = \Delta G^{\circ} + RT\ln Q \) will increase because ln Q is negative and the negative sign gets absorbed which increases the value of ΔG. #Step 2: Analyzing Reaction (b)#
02

Reaction (b):

\(2 \mathrm{HBr}(\mathrm{g}) \longrightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{Br}_{2}(g)\). For this reaction, the reaction quotient \(Q\) can be expressed as: \[Q = \frac{[\mathrm{H}_{2}][\mathrm{Br}_{2}]}{[\mathrm{HBr}]^{2}}\] If we increase the partial pressure of \(\mathrm{H}_{2}\), the value of \(Q\) will increase, because it is in the numerator of the expression for \(Q\). Thus, in this case, \( \Delta G = \Delta G^{\circ} + RT\ln Q \) will decrease since ln Q is positive. #Step 3: Analyzing Reaction (c)#
03

Reaction (c):

\(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})\). For this reaction, the reaction quotient \(Q\) can be expressed as: \[Q = \frac{[\mathrm{C}_{2} \mathrm{H}_{6}]}{[\mathrm{H}_{2}]^{2}[\mathrm{C}_{2} \mathrm{H}_{2}]}\] As we increase the partial pressure of \(\mathrm{H}_{2}\), the value of \(Q\) will decrease, as it is in the denominator of the expression for \(Q\). Therefore, in this case, \( \Delta G = \Delta G^{\circ} + RT\ln Q \) will increase since ln Q becomes more negative. #Summary# 1. For reaction (a), \(\Delta G\) increases when the partial pressure of \(\mathrm{H}_{2}\) is increased. 2. For reaction (b), \(\Delta G\) decreases when the partial pressure of \(\mathrm{H}_{2}\) is increased. 3. For reaction (c), \(\Delta G\) increases when the partial pressure of \(\mathrm{H}_{2}\) is increased.

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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 pivotal concept in chemical thermodynamics that indicates the maximum amount of work a system can perform at constant temperature and pressure. When we talk about chemical reactions, \( \Delta G \), helps determine whether a process is spontaneously favorable or not. The sign of \( \Delta G \) tells us which direction a reaction naturally tends to proceed: if \( \Delta G < 0 \), the reaction is spontaneous; if \( \Delta G > 0 \), the reaction is nonspontaneous. Reactions aim to reach a state where \( \Delta G = 0 \), known as equilibrium.

For example, by increasing the partial pressure of \( \mathrm{H}_2 \) in the given reactions, the Gibbs Free Energy changes depending on reaction stoichiometry and the reaction quotient, which reflects how far the system is from equilibrium. This analysis gives invaluable insight into the direction and extent of a chemical reaction under given conditions. Understanding how \( \Delta G \) changes with the conditions helps in designing and controlling processes in fields ranging from industrial chemistry to biochemistry.
Reaction Quotient
The Reaction Quotient, denoted as \( Q \), captures the relative concentrations of reactants and products at any point during a reaction that has not yet reached equilibrium. Essentially, it is a 'snapshot' of a reaction's progress, comparing the current state to the balanced equation of the reaction. The formula for \( Q \), depends on the reaction, but it generally follows a pattern where the concentrations of the products are divided by the concentrations of the reactants, with each raised to the power of their stoichiometric coefficients.

In our exercise, increasing the partial pressure of one reactant, \( \mathrm{H}_2 \), has varying effects on \( Q \), impacting \( \Delta G \) correspondingly. By manipulating \( Q \) through changing conditions, such as pressure or concentration, we can predict and control whether a reaction will proceed forward or reverse, which is crucial for industrial applications, such as in the synthesis of ammonia from nitrogen and hydrogen gases.
Chemical Equilibrium
Chemical Equilibrium occurs when the rates of the forward and reverse reactions are equal, leading to no net change in the amounts of reactants and products. At equilibrium, \( Q \) becomes equal to the equilibrium constant \( K \). When \( Q = K \), \( \Delta G \) is zero, indicating that the system is at maximum stability and no additional work can be extracted from it.

The examples provided underscore that as the system shifts in response to changes in conditions, such as pressure, it does so in a manner aiming to re-establish equilibrium—a central theme of reaction predictable chemistry. By studying how these shifts affect \( \Delta G \) and \( Q \), we gain a better understanding of chemical processes, control reaction outcomes, and optimize conditions for desired product formation.
Le Chatelier's Principle
Le Chatelier's Principle provides a qualitative means to predict how a system at equilibrium responds to external changes—it will adjust to minimize the effect of the change, effectively shifting the equilibrium position. An increase in pressure, for example, will shift the reaction toward the side with fewer moles of gas, if there are unequal moles on either side of the reaction equation.

In our scenario, increasing the partial pressure of \( \mathrm{H}_2 \) triggers shifts in the reaction to partially counteract this change. For reaction (a), the reaction moves towards the reactants because there are more moles of \( \mathrm{H}_2 \) on the reactant side. Reaction (b) is unique as adding more \( \mathrm{H}_2 \) shifts the reaction towards the products. Le Chatelier's Principle is a powerful tool allowing chemists to predict the direction of the shift in an equilibrium in response to changes in concentration, pressure, or temperature, thereby optimizing yield and energy efficiency in chemical production.

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Most popular questions from this chapter

For the isothermal expansion of a gas into a vacuum, \(\Delta E=0, q=0\), and \(w=0\). (a) Is this a spontaneous

An ice cube with a mass of \(20 \mathrm{~g}\) at \(-20^{\circ} \mathrm{C}\) (typical freezer temperature) is dropped into a cup that holds \(500 \mathrm{~mL}\) of hot water, initially at \(83^{\circ} \mathrm{C}\). What is the final temperature in the cup? The density of liquid water is \(1.00 \mathrm{~g} / \mathrm{mL}\), the specific heat capacity of ice is \(2.03 \mathrm{~J} / \mathrm{g}-\mathrm{C}\); the specific heat capacity of liquid water is \(4.184 \mathrm{~J} / \mathrm{g}-\mathrm{C}\), the enthalpy of fusion of water is \(6.01 \mathrm{~kJ} / \mathrm{mol}\).

The reaction $$ \mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 3 \mathrm{~S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ is the basis of a suggested method for removal of \(\mathrm{SO}_{2}\) from power-plant stack gases. The standard free energy of each substance is given in Appendix C. (a) What is the equilibrium constant for the reaction at \(298 \mathrm{~K}\) ? (b) In principle, is this reaction a feasible method of removing \(\mathrm{SO}_{2}\) ? (c) If \(\mathrm{P}_{\mathrm{so}_{2}}=\mathrm{P}_{\mathrm{A}_{2} \mathrm{~s}}\) and the vapor pressure of water is 25 torr, calculate the equilibrium \(\mathrm{SO}_{2}\) pressure in the system at \(298 \mathrm{~K}\). (d) Would you expect the process to be more or less effective at higher temperatures?

Using data from Appendix \(C\), calculate the change in Gibbs free energy for each of the following reactions. In each case indicate whether the reaction is spontaneous at \(298 \mathrm{~K}\) under standard conditions. (a) \(2 \mathrm{Ag}(s)+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{AgCl}(s)\) (b) \(\mathrm{P}_{4} \mathrm{O}_{4}(s)+16 \mathrm{H}_{2}(g) \longrightarrow 4 \mathrm{PH}_{3}(g)+10 \mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{CH}_{4}(g)+4 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{CF}_{4}(g)+4 \mathrm{HF}(g)\) (d) \(2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\)

A certain reaction has \(\Delta H^{\circ}=+23.7 \mathrm{~kJ}\) and \(\Delta 5^{\circ}=+52.4 \mathrm{~J} / \mathrm{K}\). (a) Is the reaction exothermic or endothermic? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (c) Calculate \(\Delta G^{\circ}\) for the reaction at \(298 \mathrm{~K}\). (d) Is the reaction spontaneous at \(298 \mathrm{~K}\) under standard conditions?

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