(a) Calculate the reaction Gibbs free energy of \(\mathrm{I}_{2}(\mathrm{~g}) \rightarrow\) \(2 \mathrm{I}(\mathrm{g})\) at \(1200 . \mathrm{K}(K=6.8)\) when the partial pressures of \(\mathrm{I}_{2}\) and \(\mathrm{I}\) are \(0.13\) bar and \(0.98\) bar, respectively. (b) What is the spontaneous direction of the reaction? Explain briefly.

Short Answer

Expert verified
The Gibbs free energy change \( \Delta G \) for the reaction at the given conditions can be calculated using the reaction quotient \( Q \) and the equilibrium constant \( K \) according to the equation \( \Delta G = \Delta G^{\circ} + RT\ln\left(\frac{Q}{K}\right) \). The sign of \( \Delta G \) then determines the spontaneous direction of the reaction.

Step by step solution

01

Write the expression for reaction Gibbs free energy change

Determine the reaction Gibbs free energy \( \Delta G \) using the equation \( \Delta G = \Delta G^{\circ} + RT\ln\left(\frac{Q}{K}\right) \), where \( \Delta G^{\circ} \) is the standard Gibbs free energy change, \( R \) is the gas constant (8.314 J/mol*K), \( T \) is the temperature in Kelvin, \( Q \) is the reaction quotient, and \( K \) is the equilibrium constant.
02

Calculate the reaction quotient (Q)

Calculate the reaction quotient \( Q \) using the partial pressures of the products and reactants. For the reaction \( \mathrm{I}_2(g) \rightarrow 2\mathrm{I}(g) \), \( Q \) is given by \( Q = \frac{{p_{\mathrm{I}}^2}}{{p_{\mathrm{I}_2}}} \), where \( p_{\mathrm{I}} \) is the partial pressure of \( \mathrm{I} \) and \( p_{\mathrm{I}_2} \) is the partial pressure of \( \mathrm{I}_2 \) .
03

Insert the given values and calculate \( \Delta G \)

Plug in the values for \( R \) (8.314 J/mol*K), \( T \) (1200 K), \( Q \) (calculated in Step 2), and \( K \) (6.8) into the \( \Delta G \) equation and calculate the Gibbs free energy change for the reaction.
04

Determine the sign of \( \Delta G \) and the spontaneity of the reaction

The sign of \( \Delta G \) indicates the spontaneity of the reaction. If \( \Delta G \) is negative, the reaction is spontaneous in the forward direction. If positive, the reaction is non-spontaneous in the forward direction.
05

Answer part (b) discussing the spontaneous direction of the reaction

Based on the sign of \( \Delta G \) determined in Step 4, conclude the spontaneous direction of the reaction.

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is the state of a reversible chemical reaction where the rates of the forward and reverse reactions are equal, leading to no overall change in the concentrations of reactants and products over time. It's a dynamic balance, not a static one, as molecules continuously react, converting in both directions.

At equilibrium, the ratio of the concentrations of products to reactants remains constant, which is numerically expressed by the equilibrium constant, denoted as 'K'. It's important to note that different reactions at the same temperature will have unique equilibrium constants. In the exercise, the equilibrium constant, 'K', is given as 6.8 at 1200 K for the reaction involving iodine. The concept of equilibrium is essential when talking about reaction spontaneity and the reaction quotient since it serves as a point of reference.
Reaction Spontaneity
Reaction spontaneity refers to whether a chemical reaction would occur without any additional input of energy. This is different from the rate of the reaction - spontaneous reactions can still be slow. The spontaneity of a reaction can be understood by considering Gibbs free energy, symbolized as \( \Delta G \).

A spontaneous reaction is indicated by a negative \( \Delta G \) value, meaning the reaction can proceed in the forward direction without external work. On the other hand, a positive \( \Delta G \) suggests that the reaction is not spontaneous and will not occur under the existing conditions without an input of energy.

In the textbook example, by calculating the Gibbs free energy and observing its sign, students can determine whether the reaction of \( \mathrm{I}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{I}(\mathrm{g}) \) is spontaneous at the given temperature and partial pressures. This directly relates to the fundamental concepts of thermodynamics and is an important factor for chemical reactions and processes.
Reaction Quotient
The reaction quotient (Q) is a measure that compares the relative amounts of products and reactants present during a reaction at any point in time. It is calculated in the same way as the equilibrium constant (K), but it is not limited to just the conditions at chemical equilibrium.

The value of Q provides a snapshot of a reaction's progress and offers insight into which direction the reaction will proceed to reach equilibrium. If \( Q < K \) the reaction will move forward towards the products to reach equilibrium. If \( Q > K \) the reaction will proceed in the reverse direction, converting products back into reactants.

In the step-by-step exercise provided, by calculating the reaction quotient (Q) using the given partial pressures, and comparing it with the equilibrium constant (K), students can assess the status of the reaction regarding its equilibrium position. This analysis supports the determination of the Gibbs free energy change (\(\Delta G\)) and ultimately the spontaneity of the reaction.

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Most popular questions from this chapter

Write the reaction quotient \(Q\) for (a) \(2 \mathrm{BCl}_{3}\) (g) \(+2 \mathrm{Hg}(\mathrm{l}) \rightarrow \mathrm{B}_{2} \mathrm{Cl}_{4}\) (s) \(+\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) (s) (b) \(\mathrm{P}_{4} \mathrm{~S}_{10}\) (s) \(+16 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 4 \mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})+10 \mathrm{H}_{2} \mathrm{~S}(\mathrm{aq})\) (c) \(\mathrm{Br}_{2}(\mathrm{~g})+3 \mathrm{~F}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{BrF}_{3}(\mathrm{~g})\)

For the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) at \(400 . \mathrm{K}\), \(K=41\). Find the value of \(K\) for each of the following reactions at the same temperature: (a) \(2 \mathrm{NH}_{3}\) (g) \(\rightleftharpoons \mathrm{N}_{2}\) (g) \(+3 \mathrm{H}_{2}\) (g) (b) \(\frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})\) (c) \(2 \mathrm{~N}_{2}(\mathrm{~g})+6 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NH}_{3}(\mathrm{~g})\)

At 2500 . \(K\), the equilibrium constant is \(K_{c}=20\). for the reaction \(\mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{ClF}(\mathrm{g})\). An analysis of a reaction vessel at 2500 . \(\mathrm{K}\) revealed the presence of \(0.18 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{Cl}_{2}\), \(0.31 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~F}_{2}\), and \(0.92 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{ClF}\). Will \(\mathrm{ClF}\) tend to form or to decompose as the reaction proceeds toward equilibrium?

When \(0.0172 \mathrm{~mol} \mathrm{HI}\) is heated to \(500 . \mathrm{K}\) in a \(2.00-\mathrm{L}\) sealed container, the resulting equilibrium mixture contains \(1.90 \mathrm{~g}\) of HI. Calculate \(K\) for the decomposition reaction \(2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons\) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g})\)

Consider the equilibrium \(3 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})+\) \(6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\). (a) What happens to the partial pressure of \(\mathrm{NH}_{3}\) when the partial pressure of \(\mathrm{NO}\) is increased? (b) Does the partial pressure of \(\mathrm{O}_{2}\) decrease when the partial pressure of \(\mathrm{NH}_{3}\) is decreased?

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