/*! 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 45 The standard free energy change,... [FREE SOLUTION] | 91影视

91影视

The standard free energy change, \(\Delta_{\mathrm{r}} G^{\circ},\) for the formation of \(\mathrm{NO}(\mathrm{g})\) from its elements is \(+86.58 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn}\) at \(25^{\circ} \mathrm{C} .\) Calculate \(K_{\mathrm{p}}\) at this temperature for the equilibrium $$1 / 2 \mathrm{N}_{2}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g})$$ Comment on the sign of \(\Delta_{\mathrm{r}} G^{\circ}\) and the magnitude of \(K_{\mathrm{p}}.\)

Short Answer

Expert verified
Kp at 298.15 K is approximately 7.1 脳 10鈦宦光伓. The positive 螖rG掳 indicates non-spontaneity, and the small Kp suggests reactants predominate.

Step by step solution

01

Understand the Relationship Between 螖rG掳 and Kp

The relationship between the standard free energy change of a reaction, 螖rG掳, and the equilibrium constant, K, can be determined using the equation \( \Delta_{\mathrm{r}} G^{\circ} = -RT \ln K \), where R is the universal gas constant (8.314 J/mol路K) and T is the temperature in Kelvin.
02

Convert Temperature to Kelvin

Since the temperature is given in Celsius, convert it to Kelvin using the formula \( T(K) = T(掳C) + 273.15 \). Thus, \( 25^{\circ} \mathrm{C} \) becomes \( 298.15 \text{ K} \).
03

Solve for Kp Using 螖rG掳

Substitute the given 螖rG掳 and the converted temperature into the equation: \( 86580 \text{ J/mol} = -(8.314 \text{ J/mol路K})(298.15 \text{ K}) \ln K_p \). Rearrange to find \( \ln K_p \) and solve for \( K_p \).
04

Calculate ln Kp

Rearrange the equation: \( \ln K_p = \frac{-86580}{8.314 \times 298.15} \approx -34.92 \).
05

Calculate Kp

Take the exponential of both sides to solve for \( K_p \): \( K_p = e^{-34.92} \approx 7.1 \times 10^{-16} \).
06

Interpret the Sign of 螖rG掳 and Magnitude of Kp

Given that \( \Delta_{\mathrm{r}} G^{\circ} \) is positive, this indicates that the formation of NO is not spontaneous under standard conditions. Additionally, the very small magnitude of \( K_p \) (7.1 脳 10鈦宦光伓) suggests that at equilibrium, the concentrations of reactants dominate over products, indicating a far shift to the left.

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 91影视!

Key Concepts

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

Free Energy Change
Free energy change, denoted as \( \Delta_{\text{r}} G^{\circ} \), is an important concept in thermodynamics and chemistry that describes the amount of energy available to do work during a chemical reaction. In the context of a reaction, if \( \Delta_{\text{r}} G^{\circ} \) is negative, the reaction can proceed spontaneously under standard conditions.However, if it is positive, like in the formation of NO from N鈧 and O鈧 with a value of 86.58 kJ/mol, the reaction is non-spontaneous.
The free energy change is a central measure used to determine both the spontaneity and feasibility of reactions:
  • A negative \( \Delta_{\text{r}} G^{\circ} \) means the process releases energy, spontaneously moving forward.
  • A positive \( \Delta_{\text{r}} G^{\circ} \) indicates energy must be added for the reaction to occur.
The relationship between \( \Delta_{\text{r}} G^{\circ} \) and the equilibrium constant\( K \) is given by the formula \( \Delta_{\text{r}} G^{\circ} = -RT \ln K \), where \( R \) is the universal gas constant and \( T \) is the temperature in Kelvin.
Equilibrium Equation
Equilibrium equations express the state of a chemical reaction where the rates of the forward and backward reactions are equal. This means the concentrations of reactants and products remain constant over time. Equilibrium can be represented by a constant known as the equilibrium constant \( K \).
The equilibrium constant is derived from the ratio of the concentrations of products to reactants, each raised to the power of their stoichiometric coefficients. For the given reaction:\[ \frac{1}{2} \text{N}_2(\text{g}) + \frac{1}{2} \text{O}_2(\text{g}) \rightleftharpoons \text{NO}(\text{g}) \]the constant \( K_p \) can be determined using the equation:\( \Delta_{\text{r}} G^{\circ} = -RT \ln K_p \).
In this exercise, \( \Delta_{\text{r}} G^{\circ} \) was calculated as +86.58 kJ/mol, indicating a non-spontaneous reaction, and leading to an exceptionally low \( K_p \) value of approximately \( 7.1 \times 10^{-16} \). This signifies that at equilibrium, the reactants are much more prevalent than the products, indicating the equilibrium strongly favors the reactants.
Reaction Thermodynamics
Thermodynamics is the study of energy transformations in reaction processes. It provides insights into whether chemical reactions will occur spontaneously and to what extent. Using the principles of reaction thermodynamics, one can determine the spontaneity and equilibrium position of a reaction.
The key thermodynamic parameters are:
  • Enthalpy (\( \Delta H \)): The total energy of a system. It is not directly addressed in this exercise, but it plays a crucial role in conjunction with entropy to determine spontaneity.
  • Entropy (\( \Delta S \)): The degree of randomness or disorder in a system. It influences the free energy change, and hence the equilibrium of a reaction.
  • Free Energy (\( \Delta G \)): Combines enthalpy and entropy changes to assess whether a process can be spontaneous.
Through the equation-related energy changes and equilibrium constant, reacting systems can be evaluated for feasibility and direction.

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

Iodine, I \(_{2}\), dissolves readily in carbon tetrachloride. For this process, \(\Delta H^{\circ}=0 \mathrm{kJ} / \mathrm{mol}\). $$\mathrm{I}_{2}(\mathrm{s}) \rightarrow \mathrm{I}_{2}\left(\text { in } \mathrm{CCl}_{4} \text { solution }\right)$$ What is the sign of \(\Delta_{r} G^{\circ} ?\) Is the dissolving process entropy-driven or enthalpy-driven? Explain briefly.

Indicate which of the following processes are reversible. (a) Nitrogen and oxygen gases diffuse to give a homogeneous mixture. (b) Ice sublimes at \(-5^{\circ} \mathrm{C}\) and 1.0 atm. (c) Energy as heat is transferred to the surroundings from a mixture of ice and water at \(0^{\circ} \mathrm{C}\) causing more ice to form. (d) Bromine evaporates and the gaseous molecules diffuse into the atmosphere.

Identify the following processes as either spontaneous or not spontaneous. (a) Liquid water turns to ice when placed in a freezer at \(-5^{\circ} \mathrm{C}\) (b) Nitrogen gas is compressed to one half its original volume. (c) Sodium reacts with water forming \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{NaOH}(\mathrm{aq})\) (d) Slightly soluble \(\operatorname{CaSO}_{4}\left(K_{\mathrm{sp}}=4.5 \times 10^{-5}\right)\) dis- solves in water to form a saturated solution.

Sodium reacts violently with water according to the equation $$\mathrm{Na}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{NaOH}(\mathrm{aq})+1 / 2 \mathrm{H}_{2}(\mathrm{g})$$Without doing calculations, predict the signs of \(\Delta_{\mathrm{r}} H^{\circ}\) and \(\Delta_{\mathrm{r}} S^{\circ}\) for the reaction. Verify your prediction with a calculation.

Calculate the entropy change, \(\Delta_{r} S^{\circ},\) for the vaporization of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH},\) at its normal boiling point, \(78.0^{\circ} \mathrm{C} .\) The enthalpy of vaporization of ethanol is \(39.3 \mathrm{kJ} / \mathrm{mol}\).

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.