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Chapter 14: Problem 69

At \(298 \mathrm{~K}, K_{\mathrm{p}}\) is \(1.6 \times 10^{-6}\) for the reaction \(2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) .\) What is the value of \(K_{c}\) at this temperature?

Short Answer

Expert verified
At 298 K, \( K_c \) is approximately \( 6.54 \times 10^{-8} \).

Step by step solution

01

Understanding the Relationship Between Kp and Kc

The equilibrium constants for gases in terms of pressure \( K_p \) and concentration \( K_c \) are related by the equation \( K_c = \frac{K_p}{(RT)^{\Delta n}} \), where \( R \) is the gas constant, \( T \) is the temperature in Kelvin, and \( \Delta n \) is the change in the number of moles of gas.
02

Identifying Parameters and Constants

From the problem, we have \( K_{p} = 1.6 \times 10^{-6} \) at \( T = 298 \mathrm{~K} \). The gas constant \( R \) is \( 0.0821 \frac{L\cdot atm}{K\cdot mol} \). We need to calculate \( \Delta n \), the change in moles of gas in the reaction.
03

Calculating Change in Moles (螖n)

For the reaction \( 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g) + \mathrm{Cl}_{2}(g) \), the change in moles \( \Delta n = (2 + 1) - 2 = 1 \).
04

Substituting Values Into the Kc Formula

Substitute \( K_p = 1.6 \times 10^{-6} \), \( R = 0.0821 \), \( T = 298 \), and \( \Delta n = 1 \) into the formula: \[ K_c = \frac{1.6 \times 10^{-6}}{(0.0821 \times 298)^1} \]
05

Performing the Calculation

Calculate the value inside the parenthesis first: \( 0.0821 \times 298 = 24.4718 \). Now calculate \( K_c = \frac{1.6 \times 10^{-6}}{24.4718} \approx 6.54 \times 10^{-8} \).
06

Concluding the Solution

The value of \( K_c \) at \( 298 \) K is approximately \( 6.54 \times 10^{-8} \).

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Key Concepts

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

Kp and Kc relationship
The equilibrium constants, represented as \( K_p \) and \( K_c \), are essential for understanding chemical reactions, especially gases. Both constants help in describing the extent of a reaction at equilibrium. Here's how they differ:
  • \( K_p \) is the equilibrium constant expressed in terms of the partial pressures of gases.
  • \( K_c \) is the equilibrium constant expressed in terms of concentrations (molarity) of gases.
These constants are interrelated by the equation:\[ K_c = \frac{K_p}{(RT)^{\Delta n}} \]
  • \( R \) is the ideal gas constant.
  • \( T \) is the temperature in Kelvin.
  • \( \Delta n \) is the change in moles of gas (products minus reactants).
This relationship is critical because it allows us to convert between conditions of pressure and concentration, which is especially useful under different experimental setups.
gas constant
The gas constant \( R \) plays a vital role in the relationship between \( K_p \) and \( K_c \). It is a universal constant that links various physical properties of gases in the ideal gas law \( PV = nRT \), where:
  • \( P \) is pressure, \( V \) is volume.
  • \( n \) is the number of moles.
  • \( T \) is temperature in Kelvin.
  • \( R \) is the ideal gas constant, with a common value of \( 0.0821 \frac{L\cdot atm}{K\cdot mol} \).
The role of \( R \) in equilibrium calculations is to adjust for the units of pressure and concentration.Whenever \( K_p \) values are converted to \( K_c \) or vice versa, \( R \) is crucial for ensuring the units are consistent.This is especially important in problems like predicting the behavior of gaseous reactions at particular temperatures.
reaction stoichiometry
Stoichiometry of a chemical equation describes how moles of reactants combine to form products. It is essential in calculating \( \Delta n \) (change in moles), a key parameter in the \( K_p \) and \( K_c \) relationship.For example, consider the reaction:\[ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g) + \mathrm{Cl}_2(g) \]
  • Moles of products: 2 (NO) + 1 (\( \mathrm{Cl}_2 \)) = 3 moles
  • Moles of reactants: 2 (NOCl) = 2 moles
  • \( \Delta n \) = 3 - 2 = 1
The change in moles \( \Delta n \) is crucial for converting \( K_p \) to \( K_c \).Accurate stoichiometry ensures precise calculations and helps predict the direction and extent of chemical reactions.
thermodynamics
Thermodynamics explores how energy changes affect chemical reactions.One key aspect is how temperature influences equilibrium constants, such as \( K_p \) and \( K_c \).
  • Higher temperatures can increase reaction rates but may shift equilibrium constants depending on whether a reaction is endothermic or exothermic.
  • Exothermic reactions release heat and can have decreased equilibrium constants with temperature rise.
  • Endothermic reactions absorb heat and may have increased equilibrium constants with higher temperatures.
Understanding these influences is vital for predicting how a system will respond to changes in temperature, which is crucial when working with equilibrium in reactions.By carefully controlling temperature, chemists can manipulate reactions to optimize yield or speed, directly linking thermodynamic principles to practical applications.

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Most popular questions from this chapter

The decomposition of solid ammonium carbamate, \(\left(\mathrm{NH}_{4}\right)\) \(\left(\mathrm{NH}_{2} \mathrm{CO}_{2}\right)\), to gaseous ammonia and carbon dioxide is an endothermic reaction. $$ \left(\mathrm{NH}_{4}\right)\left(\mathrm{NH}_{2} \mathrm{CO}_{2}\right)(s) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) $$ (a) When solid \(\left(\mathrm{NH}_{4}\right)\left(\mathrm{NH}_{2} \mathrm{CO}_{2}\right)\) is introduced into an evacuated flask at \(25^{\circ} \mathrm{C}\), the total pressure of gas at equilibrium is \(0.116 \mathrm{~atm}\). What is the value of \(K_{\mathrm{p}}\) at \(25^{\circ} \mathrm{C} ?\) (b) Given that the decomposition reaction is at equilibrium, how would the following changes affect the total quantity of \(\mathrm{NH}_{3}\) in the flask once equilibrium is re-established? (i) Adding \(\mathrm{CO}_{2}\) (ii) Adding \(\left(\mathrm{NH}_{4}\right)\left(\mathrm{NH}_{2} \mathrm{CO}_{2}\right)\) (iii) Removing \(\mathrm{CO}_{2}\) (iv) Increasing the total volume (v) Adding neon (vi) Increasing the temperature

For each of the following equilibria, use Le Ch芒telier's principle to predict the direction of reaction when the volume is increased. (a) \(\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2}(g)\) (b) \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g)\) (c) \(2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g)\)

Halogen lamps are ordinary tungsten filament lamps in which the lamp bulb contains a small amount of a halogen (often bromine). At the high temperatures of the lamp, the halogens dissociate and exist as single atoms. (a) In an ordinary tungsten lamp, the hot tungsten filament is constantly evaporating and the tungsten condenses on the relatively cool walls of the bulb. In a Br-containing halogen lamp, the tungsten reacts with the \(\mathrm{Br}\) atoms to give gaseous \(\mathrm{WBr}_{4}:\) $$ \mathrm{W}(s)+4 \mathrm{Br}(g) \rightleftharpoons \mathrm{WBr}_{4}(g) $$ At the walls of the lamp, where the temperature is about \(900 \mathrm{~K}\), this reaction has an equilibrium constant \(K_{\mathrm{p}}\) of about \(100 .\) If the equilibrium pressure of \(\mathrm{Br}(g)\) is \(0.010 \mathrm{~atm}\), what is the equilibrium pressure of \(\mathrm{WBr}_{4}(g)\) near the walls of the bulb? (b) Near the tungsten filament, where the temperature is about \(2800 \mathrm{~K}\), the reaction in part (a) has a \(K_{p}\) value of about \(5.0\). Is the reaction exothermic or endothermic? (c) When the \(\mathrm{WBr}_{4}(g)\) diffuses back toward the filament, it decomposes, depositing tungsten back onto the filament. Show quantitatively that the pressure of \(\mathrm{WBr}_{4}\) from part (a) will cause the reaction in part (a) to go in reverse direction at \(2800 \mathrm{~K}\). [The pressure of \(\mathrm{Br}(\mathrm{g})\) is still \(0.010 \mathrm{~atm}\). ] Thus, tungsten is continually recycled from the walls of the bulb back to the filament, allowing the bulb to last longer and burn brighter.

The equilibrium constant \(K_{\mathrm{p}}\) for the gas-phase thermal decomposition of tert-butyl chloride is \(3.45\) at \(500 \mathrm{~K}\) : $$ \left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}(g) \rightleftharpoons\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CH}_{2}(g)+\mathrm{HCl}(g) $$ (a) Calculate the value of \(K_{c}\) at \(500 \mathrm{~K}\). (b) Calculate the molar concentrations of reactants and products in an equilibrium mixture obtained by heating \(1.00 \mathrm{~mol}\) of tert-butyl chloride in a \(5.00 \mathrm{~L}\) vessel at \(500 \mathrm{~K}\). (c) A mixture of isobutylene \((0.400\) atm partial pressure at \(500 \mathrm{~K}\) ) and \(\mathrm{HCl}(0.600 \mathrm{~atm}\) partial pressure at \(500 \mathrm{~K}\) ) is allowed to reach equilibrium at \(500 \mathrm{~K}\). What are the equilibrium partial pressures of tert-butyl chloride, isobutylene, and \(\mathrm{HCl} ?\)

Identify the true statement about the rate of the forward and reverse reaction once a reaction has reached equilibrium. (a) The rate of the forward reaction and the reverse reaction is zero. (b) The rate of the forward reaction is greater than the rate of the reverse reaction. (c) The rate of the reverse reaction is greater than the rate of the forward reaction. (d) The rate of the forward reaction is equal to the rate of the reverse reaction.
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