Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Given that \(\mathrm{k}=8.85\) at \(298^{\circ} \mathrm{K}\) and \(\mathrm{k}=.0792\) at \(373^{\circ} \mathrm{K}\), calculate the \(\Delta \mathrm{H}^{\circ}\) for the reaction of the dimerization of \(\mathrm{NO}_{2}\) to \(\mathrm{N}_{2} \mathrm{O}_{4}\). Namely, \(2 \mathrm{NO}_{2}(\mathrm{~g}) \rightleftarrows \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\).

Short Answer

Expert verified
The standard enthalpy change for the dimerization of NO₂ to N₂O₄, given the equilibrium constants at two temperatures, is calculated using the Van't Hoff equation. By inserting the given values (k1 = 8.85 at T1 = 298 K and k2 = 0.0792 at T2 = 373 K) and solving for ΔH, we find the standard enthalpy change to be approximately \(-57.4 \, \mathrm{kJ/mol}\).

Step by step solution

01

Write down the Van't Hoff equation

The Van't Hoff equation relates the equilibrium constant k, temperature T, and the standard enthalpy change \(\Delta \mathrm{H}^{\circ}\) for the reaction: \[\frac{d(\ln k)}{dT} = \frac{\Delta \mathrm{H}^{\circ}}{R T^{2}}\]
02

Integrate the equation

For two points, k1 at T1 and k2 at T2, integrate the equation by considering d( ln k) on the left and \( \Delta H^{\circ}/RT^{2}\) dT on the right: \[ \int_{ln k_1}^{ln k_2} d(\ln k) = \int_{T_1}^{T_2}\frac{ \Delta \mathrm{H}^{\circ} }{R T^{2}} dT \]
03

Calculate the result of integration

After integrating, we'll have: \[(\ln k_2 - \ln k_1) = \frac{\Delta \mathrm{H}^{\circ}}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)\] Now, we can insert the given values for k1, k2, T1, and T2 and solve the equation for ΔH.
04

Insert the given values

Given values: k1 = 8.85 at T1 = 298 K and k2 = 0.0792 at T2 = 373 K. Also, R (gas constant) = 8.314 J/(mol·K). So, we have: \[(\ln 0.0792 - \ln 8.85) = \frac{\Delta \mathrm{H}^{\circ}}{8.314} \left( \frac{1}{298} - \frac{1}{373} \right)\]
05

Solve for ΔH

Calculate ΔH from the equation: \[\Delta \mathrm{H}^{\circ} = \frac{(\ln 0.0792 - \ln 8.85) \cdot 8.314}{\frac{1}{298} - \frac{1}{373}}\] \[\Delta \mathrm{H}^{\circ} = -57447.59 \, \mathrm{J/mol}\] The standard enthalpy change for the dimerization of NO₂ to N₂O₄ is approximately -57.4 kJ/mol.

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.

Equilibrium Constant
Understanding the equilibrium constant is essential for grasping the heart of many chemical reactions. In a balanced chemical equation, the equilibrium constant, denoted as K, gives us a numeric value that represents the ratio of the concentrations of products to reactants at equilibrium. It's a snapshot of the reaction's state when the forward and reverse reaction speeds are equal, meaning no net change is observed in concentration over time. This constant is crucial because it can tell us the extent to which a reaction will proceed under certain conditions. A high K value indicates a reaction heavily favored to produce products, while a low K suggests a balance leaning towards reactants.

In the Van't Hoff equation, K is represented by 'k,' and its changes with temperature provide insights into thermodynamics and chemical kinetics of a reaction. Still, it is important to note that the 'k' used in the Van't Hoff equation should not be confused with the rate constant of a reaction, which is a different concept altogether.
Standard Enthalpy Change
The term standard enthalpy change, \( \Delta \mathrm{H}^\circ \), is a thermodynamic concept that measures the total heat change during a reaction under standard conditions (1 atm and often at 298 K). It's one of the most significant indicators of how much energy is absorbed or released during a chemical reaction. A negative \( \Delta \mathrm{H}^\circ \) signifies an exothermic process where heat is given off to the surroundings, whilst a positive value indicates an endothermic one, where the reaction absorbs heat from its environment.

In the context of the Van't Hoff equation, knowing the \( \Delta \mathrm{H}^\circ \) helps us predict how the equilibrium constant will shift with changes in temperature, enabling chemists to tailor conditions for optimal yield of desired products.
Chemical Kinetics
Chemical kinetics is the study of the rates of chemical processes and the factors that affect these rates. It examines how various variables like concentration, temperature, and catalysts can alter the speed at which reactants convert into products. This field is crucial for developing and optimizing new reactions in industry and research, as it helps predict how long a reaction will take to reach equilibrium or to produce a desired amount of product.

While the equilibrium constant is tied to the thermodynamic aspect, which doesn't involve time, kinetics bridges time with reaction rates. The Van't Hoff equation, primarily used in thermodynamics, provides a link to kinetics by showing how the equilibrium state of a reaction changes with temperature, indirectly hinting at the rate changes due to temperature shifts.
Thermodynamics
Thermodynamics is the branch of physical science that deals with the relations between heat and other forms of energy. In the world of chemistry, it helps us understand the energy changes that accompany chemical reactions and physical transformations. Three main laws of thermodynamics guide these relationships, contributing to concepts such as enthalpy, entropy, and Gibbs free energy.

The Van't Hoff equation provides a window into the thermodynamic properties of a chemical reaction. By studying how the equilibrium constant varies with temperature, we can deduce important information about reaction spontaneity and prevailing conditions for a reaction to occur. Crucially, it integrates the equilibrium state of a system, represented by its equilibrium constant, with the enthalpy change and temperature — fundamental components of thermodynamics.

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

Hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\), can be synthesized in two ways. The first method involves reduction of oxygen by hydrogen, $$ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}_{2}(\ell) $$ The second method involves oxidation of water: $$ 2 \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}_{2}(\ell) $$ Find the free energy of formation, \(\Delta \mathrm{G}^{\circ}\), for both processes and predict which process is more efficient for the commercial preparation of hydrogen peroxide.

The equilibrium constant and standard Gibbs free energy change for ammonia synthesis at \(400^{\circ} \mathrm{C}\) or \(673^{\circ} \mathrm{K}\) are \(1.64 \times\) \(10^{-4}\) and \(11,657 \mathrm{cal} /\) mole, respectively. The equation for this reaction is (1) \(\quad \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) Calculate the equilibrium constants and standard free energy changes for (2) \((1 / 2) \mathrm{N}_{2}(\mathrm{~g})+3 / 2 \mathrm{H}_{2}(\mathrm{~g})=\mathrm{NH}_{3}(\mathrm{~g})\) and (3) \(2 \mathrm{NH}_{3}(\mathrm{~g}) \rightarrow \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})\)

Calculate \(\Delta \mathrm{S}\) for the conversion of one mole of liquid water to vapor at \(100^{\circ} \mathrm{C}\). Heat of vaporization \(=540 \mathrm{cal} / \mathrm{g}\).

If the standard free energy of formation of HI from \(\mathrm{H}_{2}\) and \(\mathrm{I}_{2}\) at \(490^{\circ} \mathrm{C}\) is \(-12.1 \mathrm{~kJ} /\) mole of \(\mathrm{HI}\), what is the equilibrium constant for this reaction? Assume \(\mathrm{R}=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{deg}^{-1}\)

Determine the entropy change that takes place when 1 mole of ammonia (a) Passes from the liquid state to the gaseous state at its boiling point, \(-33^{\circ} \mathrm{C} ; \Delta \mathrm{H}_{\mathrm{VAP}}=5570 \mathrm{cal} / \mathrm{mole}\) (b) as a gas at \(-33^{\circ} \mathrm{C}\) comes to room temperature, \(25^{\circ} \mathrm{C}\). Assume heat capacity is constant at \(8.9\) cal/deg - mole for this range.

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