/*! 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 68 At \(298 \mathrm{~K}, K_{c}\) is... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

At \(298 \mathrm{~K}, K_{c}\) is \(2.2 \times 10^{5}\) for the reaction \(\mathrm{F}(g)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{O}_{2} \mathrm{~F}(g) .\) What is the value of \(K_{\mathrm{p}}\) at this temperature?

Short Answer

Expert verified
The value of \(K_{p}\) is approximately 8989.80.

Step by step solution

01

Understanding the Relation between Kc and Kp

To solve for \(K_{p}\), we need to understand the relationship between \(K_{c}\) and \(K_{p}\). This relationship is given by the equation: \[ K_{p} = K_{c} \left( RT \right)^{\Delta n} \] where \(R\) is the ideal gas constant \(0.0821 \text{ L atm mol}^{-1} \text{ K}^{-1}\), \(T\) is the temperature in Kelvin, and \(\Delta n\) is the change in moles of gas between products and reactants.
02

Calculate Δn for the Reaction

For the reaction \( \text{F}(g) + \text{O}_2(g) \rightleftharpoons \text{O}_2\text{F}(g) \), calculate \(\Delta n\). Count the moles of gaseous reactants and products.- Moles of gaseous reactants: 1 (F) + 1 (O2) = 2- Moles of gaseous products: 1 (O2F)Thus, \(\Delta n = 1 - 2 = -1\).
03

Apply the Formula to Find Kp

Substitute the given \(K_{c} = 2.2 \times 10^{5}\), \(R = 0.0821\), \(T = 298\), and \(\Delta n = -1\) into the equation \(K_{p} = K_{c} \left( RT \right)^{\Delta n}\). \[ K_{p} = 2.2 \times 10^{5} \left( 0.0821 \times 298 \right)^{-1} \]Calculate \(RT\):\[ RT = 0.0821 \times 298 = 24.4758 \]Then,\[ K_{p} = 2.2 \times 10^{5} \times \frac{1}{24.4758}\]
04

Compute the Final Value for Kp

Divide and multiply to solve for \(K_{p}\):\[ K_{p} = 2.2 \times 10^{5} \times \frac{1}{24.4758} \approx 8989.80 \]Thus, the value of \(K_{p}\) at 298 K is approximately \(8989.80\).

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 91Ó°ÊÓ!

Key Concepts

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

Kc and Kp Relationship
When dealing with chemical reactions involving gases, we often encounter two types of equilibrium constants: \(K_c\) and \(K_p\). The key difference between these constants lies in their expression: \(K_c\) is used for concentrations in molarity, while \(K_p\) pertains to partial pressures. To convert \(K_c\) to \(K_p\), we utilize the equation: \[K_p = K_c (RT)^{\Delta n}\] where:
  • \(R\) is the ideal gas constant \(0.0821 \text{ L atm mol}^{-1} \text{ K}^{-1}\)
  • \(T\) is the temperature in Kelvin
  • \(\Delta n\) is the change in moles of gaseous products minus reactants
This equation helps compute \(K_p\) when \(K_c\) is known, as long as the reaction occurs under conditions where ideal gas laws apply. The exponential term \((RT)^{\Delta n}\) accounts for the pressure effects due to the change in gaseous moles, ensuring accurate conversion.
Ideal Gas Constant
The ideal gas constant, represented by \(R\), is a fundamental value in the ideal gas law, establishing the link between pressure, volume, temperature, and amount of gas. In the context of equilibria where pressure is significant, \(R\) aids in translating concentration-based equilibrium constants \(K_c\) into pressure-based \(K_p\).Key points about the ideal gas constant:
  • It is commonly valued at \(0.0821 \text{ L atm mol}^{-1} \text{ K}^{-1}\).
  • \(R\) is derived from the ideal gas law equation \(PV = nRT\).
  • It ensures the compatibility of units when calculating \(K_p\) from \(K_c\) given that pressures or volumes of gases are involved.
The role of \(R\) in reactions involving gases is crucial for prediction and calculation of behavior under various conditions, especially where interconversion of different forms of the equilibrium constant is needed.
Reaction Moles Change
In chemical reactions involving gases, the change in the number of moles of gas, \(\Delta n\), plays a pivotal role in determining pressure-based equilibrium constants like \(K_p\). This value is calculated by subtracting the total moles of gaseous reactants from those of gaseous products.Steps for calculating \(\Delta n\):
  • Count the moles of all gaseous products.
  • Count the moles of all gaseous reactants.
  • Subtract the reactant moles from the product moles: \(\Delta n = \text{(moles of products)} - \text{(moles of reactants)}\).
In our given reaction, \(\text{F}(g) + \text{O}_2(g) \rightleftharpoons \text{O}_2\text{F}(g)\), \(\Delta n\) is calculated as follows:- 1 mole of \(\text{O}_2\text{F}(g)\) (product) vs. 2 moles of reactants (1 mole of \(\text{F}(g)\) + 1 mole of \(\text{O}_2(g)\)).- Therefore, \(\Delta n = 1 - 2 = -1\).Taking into account \(\Delta n\) allows us to accurately determine \(K_p\) from \(K_c\), adjusting for the volume and pressure changes happening in the gas phase.

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

Consider the sublimation of mothballs at \(27^{\circ} \mathrm{C}\) in a room having dimensions \(8.0 \mathrm{ft} \times 10.0 \mathrm{ft} \times 8.0 \mathrm{ft}\). Assume that the mothballs are pure solid naphthalene (density \(\left.1.16 \mathrm{~g} / \mathrm{cm}^{3}\right)\) and that they are spheres with a diameter of \(12.0 \mathrm{~mm}\). The equilibrium constant \(K_{c}\) for the sublimation of naphthalene is \(5.40 \times 10^{-6}\) at \(27{ }^{\circ} \mathrm{C}\). $$ \mathrm{C}_{10} \mathrm{H}_{8}(s) \rightleftharpoons \mathrm{C}_{10} \mathrm{H}_{\mathrm{8}}(g) $$ (a) When excess mothballs are present, how many gaseous naphthalene molecules are in the room at equilibrium? (b) How many mothballs are required to saturate the room with gaseous naphthalene?

The \(\mathrm{H}_{2} / \mathrm{CO}\) ratio in mixtures of carbon monoxide and hydrogen (called synthesis gas) is increased by the water-gas shift reaction \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons\) \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\), which has an equilibrium constant \(K_{\mathrm{c}}=4.24\) at \(800 \mathrm{~K}\). Calculate the equilibrium concentrations of \(\mathrm{CO}_{2}, \mathrm{H}_{2}, \mathrm{CO}\), and \(\mathrm{H}_{2} \mathrm{O}\) at \(800 \mathrm{~K}\) if only \(\mathrm{CO}\) and \(\mathrm{H}_{2} \mathrm{O}\) are present initially at concentrations of \(0.150 \mathrm{M}\).

A sample of HI (9.30 \(\times 10^{-3} \mathrm{~mol}\) ) was placed in an empty \(2.00 \mathrm{~L}\) container at \(1000 \mathrm{~K}\). After equilibrium was reached, the concentration of \(\mathrm{I}_{2}\) was \(6.29 \times 10^{-4} \mathrm{M} .\) Calculate the value of \(K_{c}\) at 1000 \(\mathrm{K}\) for the reaction \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\).

Naphthalene, a white solid used to make mothballs, has a vapor pressure of \(0.10 \mathrm{~mm} \mathrm{Hg}\) at \(27{ }^{\circ} \mathrm{C}\). Calculate the values of \(K_{\mathrm{p}}\) and \(K_{c}\) at \(27{ }^{\circ} \mathrm{C}\) for the equilibrium \(\mathrm{C}_{10} \mathrm{H}_{8}(s) \rightleftharpoons \mathrm{C}_{10} \mathrm{H}_{8}(g)\).

Ethylene glycol, used as antifreeze in automobile radiators, is manufactured by the hydration of ethylene oxide. Write the equilibrium constant expression for \(K_{c}\) \(\mathrm{O}\) \(\mathrm{H}_{2} \mathrm{C}-\mathrm{CH}_{2}(\mathrm{soln})+\mathrm{H}_{2} \mathrm{O}(\mathrm{soln}) \rightleftharpoons \mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}(\mathrm{soln})\)

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.