(a) Calculate the reaction free energy of \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{Hl}(\mathrm{g})\) at \(700 \mathrm{~K}\) when the concentrations of \(\mathrm{H}_{2}, \mathrm{I}_{2}\), and HI are \(0.026,0.33\), and \(1.84 \mathrm{~mol} \cdot \mathrm{L}^{-1}\), respectively. For this reaction, \(K_{c}=54\) at \(700 \mathrm{~K}\). (b) Indicate whether this reaction mixture is likely to form reactants, is likely to form products, or is at equilibrium.

Short Answer

Expert verified
\(\Delta G = 11.606 kJ/mol\), and since Q > \(\rK_c\), the mixture is likely to form reactants to reach equilibrium.

Step by step solution

01

Write the expression for the reaction quotient Q

To determine the reaction free energy, we first need to calculate the reaction quotient Q, which is a measure of the relative amounts of products and reactants present during a reaction at a particular point in time. The expression for Q for the given reaction is Q = [HI]^2 / ([H2][I2]), where the square brackets denote the concentration of the species in mol/L.
02

Calculate the reaction quotient Q

Using the concentrations given in the problem, we can calculate Q as follows: Q = (1.84)^2 / (0.026*0.33).
03

Evaluate Q

Plugging the concentrations into the expression for Q, we get Q = (1.84 * 1.84) / (0.026 * 0.33) = 3.3856 / 0.00858 = 394.6406.
04

Write the expression for the reaction free energy \(\Delta G\r)

The reaction free energy (\(\rDelta G\)) at non-standard conditions can be calculated from the equation \(\rDelta G = \rDelta G^\circ + RTln(Q)\), where R is the gas constant (8.314 J/mol*K), T is the temperature in kelvins, and \(\rDelta G^\circ\) is the standard free energy change. In this case, however, we can use the relationship \(\rDelta G = RTln(Q/K_c)\) because we know the reaction quotient Q and the equilibrium constant \(\rK_c\) at 700K.
05

Calculate \(\Delta G\) at 700 K

Substitute the values of Q, \(\rK_c\), R, and T into the equation to calculate \(\rDelta G\): \(\rDelta G = (8.314 \cdot 700)\ln(394.6406/54)\).
06

Evaluate \(\Delta G\)

Perform the calculations to find \(\rDelta G\): \(\rDelta G = 5829.8\ln(7.3093) = 5829.8 \cdot 1.9906 = 11605.788 J/mol\) or \(\r11605.788/1000 = 11.606 kJ/mol\), giving us the reaction free energy at 700 K.
07

Analyze the sign of \(\Delta G\) and compare Q with \(\rK_c\)

Since \(\Delta G\) is positive and Q > \(\rK_c\), the reaction mixture is likely to form reactants, meaning it will shift toward the reactants to reach equilibrium.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Chemical Equilibrium
Chemical equilibrium occurs when a chemical reaction and its reverse reaction proceed at the same rate, resulting in no net change in the concentration of reactants and products over time. This state can be visually represented as a see-saw balanced in the middle, symbolizing that while reactants are continuously converting to products, and vice versa, the amounts remain constant.

At equilibrium, the system's properties, such as concentration, color, and pressure, will appear static because the forward and reverse reactions are in perfect sync. However, this doesn't mean the reactions have ceased; they are ongoing but in exact opposition, creating a dynamic but stable system.
Reaction Quotient (Q)
The reaction quotient (Q) is a measure that tells us how far a reaction has processed toward equilibrium at a given moment. The formula for Q is based on the concentrations of the chemical species involved in the reaction at non-equilibrium conditions. For a general reaction given as \(aA + bB \rightarrow cC + dD\), the reaction quotient \(Q\) is defined as \(Q = \frac{[C]^c[D]^d}{[A]^a[B]^b}\), where the concentrations are those at the specific moment being assessed, not necessarily at equilibrium.

By calculating Q and comparing it to the equilibrium constant (Kc), we can predict the direction in which the reaction will proceed to reach equilibrium. If \(Q < K_c\), the reaction will proceed in the forward direction, forming more products. Conversely, if \(Q > K_c\), the system will shift to form more reactants.
Gibbs Free Energy (ΔG)
Gibbs free energy (\(\Delta G\)) is a thermodynamic quantity that helps us understand whether a reaction will occur spontaneously at constant pressure and temperature. It's defined as the energy difference between the enthalpy of a system and the product of its entropy and temperature (\(\Delta G = \Delta H - T\Delta S\)).

A negative \(\Delta G\) value indicates the reaction occurs spontaneously, producing more products, while a positive value suggests the reaction is non-spontaneous and favors the formation of reactants. At equilibrium, \(\Delta G\) is zero, signifying there's no free energy change, as the system is in its most stable state. By understanding \(\Delta G\), we gain insight into the energetic favorability of a chemical process and its potential to do work.
Equilibrium Constant (Kc)
The equilibrium constant (Kc) reflects the relative quantities of reactants and products at chemical equilibrium. Unlike Q, which applies at any moment, Kc specifically addresses the state where the reaction is balanced. Given for a reaction of the form \(aA + bB \rightarrow cC + dD\), Kc appears as \(K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\), where the concentrations are those at equilibrium.

Kc is a constant at a given temperature, meaning it's independent of initial concentrations. It provides valuable insight into the reaction's nature; a large Kc implies a product-favored reaction, while a small Kc indicates a reactant-favored reaction. In our exercise, a Kc of 54 at 700K suggests the formation of products is favored over reactants at equilibrium for the specific reaction.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

At \(1565 \mathrm{~K}\), the equilibrium constants for the reactions (1) \(2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) and \((2)\) \(2 \mathrm{CO}_{2}(\mathrm{~g})=2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\) are \(1.6 \times 10^{-11}\) and \(1.3 \times 10^{-10}\), respectively. (a) What is the equilibrium constant for the reaction (3) \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})=\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g})\) at that temperature? (b) Show that the manner in which equilibrium constants are calculated is consistent with the manner in which the \(\Delta G_{1}^{\circ}\) values are calculated when combining two or more equarions by determining \(\Delta G_{e}{ }^{\circ}\) for \((1)\) and \((2)\) and using those values to calculare \(\Delta G,{ }^{\circ}\) and \(K_{3}\) for reaction (3).

If \(Q=1.0\) for the reaction \(N_{2}(g)+O_{2}(g) \rightarrow\) \(2 \mathrm{NO}(\mathrm{g})\) at \(25^{\circ} \mathrm{C}\), will the reaction have a tendency to form products or reactants, or will it be at equilibrium?

A mixture consisting of \(1.000 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) and \(1.000 \mathrm{~mol} \mathrm{CO}(\mathrm{g})\) is placed in a \(10.00-\mathrm{L}\) reaction vessel at \(800 \mathrm{~K}\). At equilibrium, \(0.665 \mathrm{~mol} \mathrm{CO}_{2}(\mathrm{~g})\) is present as a result of the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})=\) \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\). What are (a) the equilibrium concentrations for all substances and (b) the value of \(K_{e}\) ?

A reaction mixture consisting of \(2.00 \mathrm{~mol} \mathrm{CO}\) and \(3.00 \mathrm{~mol} \mathrm{H}_{2}\) is placed into a \(10.0-\mathrm{L}\) reaction vessel and heated to \(1200 \mathrm{~K}\). At cquilibrium, \(0.478 \mathrm{~mol} \mathrm{CH}_{4}\) was present in the system. Determine the value of \(K_{c}\) for the reaction \(\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{~g})+\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\).

\( \mathrm{Ar} 500^{\circ} \mathrm{C}, K_{\mathrm{c}}=0.061\) for \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons\) \(2 \mathrm{NH}_{3}(\mathrm{~g}) .\) Calculate the value of \(K_{\varepsilon}\) at \(500^{\circ} \mathrm{C}\) for the following reactions. (a) \(\frac{1}{6} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \frac{1}{3} \mathrm{NH}_{3}(\mathrm{~g})\) (b) \(\mathrm{NH}_{3}(g)=\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{1}{2} \mathrm{H}_{2}(g)\) (c) \(4 \mathrm{~N}_{2}(\mathrm{~g})+12 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 8 \mathrm{NH}_{3}(\mathrm{~g})\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free