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Chapter 14: Problem 19

Under what conditions are the numerical values of \(K_{c}\) and \(K_{\mathrm{p}}\) equal?

Short Answer

Expert verified
Short Answer: The equilibrium constants Kc and Kp will be numerically equal under two conditions: (1) when the change in the number of moles of gas during the reaction is zero (螖n = 0), or (2) for non-zero 螖n, if the temperature T is chosen such that $$T = \left(\frac{1}{R}\right)^{-1/\Delta n}$$, where R is the ideal gas constant.

Step by step solution

01

Relationship between Kc and Kp

Recall that the relationship between Kc (equilibrium constant in terms of concentrations) and Kp (equilibrium constant in terms of partial pressures) can be written as follows: $$K_{\mathrm{p}} = K_{\mathrm{c}} \left(RT\right)^{\Delta n}$$ where: - R is the ideal gas constant (0.0821 L atm K鈦宦� mol鈦宦�) - T is the temperature in Kelvin - 螖n is the change in the number of moles of gas during the reaction (moles of gaseous products minus moles of gaseous reactants) Now let's analyze when these values are equal.
02

Conditions for numerical equality of Kc and Kp

In order for Kc and Kp to be numerically equal, we need the following equation to hold: $$K_{\mathrm{p}} = K_{\mathrm{c}} \Leftrightarrow K_{\mathrm{c}} \left(RT\right)^{\Delta n} = K_{\mathrm{c}}$$ The only way this equation will hold is if the term on the right side equals 1: $$\left(RT\right)^{\Delta n} = 1$$ Let's consider two possible scenarios for 螖n to make this condition true.
03

Scenario 1: 螖n = 0

If the change in the number of moles of gas during the reaction is zero (螖n = 0), then the term (RT)^螖n will equal 1: $$\left(RT\right)^{\Delta n} = \left(RT\right)^0 = 1$$ In this case, regardless of the values of R and T, Kc and Kp will be numerically equal.
04

Scenario 2: T = (RT)^{-1/螖n}

Another possibility exists if there is a non-zero 螖n, but the value of T is chosen such that the term (RT)^螖n becomes equal to 1: $$\left(RT\right)^{\Delta n} = 1 \Leftrightarrow T = \left(\frac{1}{R}\right)^{-1/\Delta n}$$ In this case, we have: $$K_{\mathrm{p}} = K_{\mathrm{c}} \left(\frac{1}{R}\right)^{-\Delta n/\Delta n}$$ $$K_{\mathrm{p}} = K_{\mathrm{c}} \times 1 = K_{\mathrm{c}}$$ So, Kc and Kp will be numerically equal.
05

Conclusion

The numerical values of Kc and Kp will be equal under the following conditions: 1. The change in the number of moles of gas during the reaction is zero (螖n = 0). 2. For non-zero 螖n, the temperature T is chosen such that $$T = \left(\frac{1}{R}\right)^{-1/\Delta n}$$, where R is the ideal gas constant.

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Key Concepts

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

Kc and Kp
The equilibrium constants \(K_c\) and \(K_p\) are fundamental in understanding chemical reactions at equilibrium. They represent different ways to express the equilibrium position of a reaction. \(K_c\) is used when dealing with concentrations of reactants and products in a solution, usually measured in mol/L. On the other hand, \(K_p\) is applicable when dealing with gases, using partial pressures measured in atm or other pressure units.

These constants are related by the equation:
  • \( K_{p} = K_{c} \left(RT\right)^{\Delta n} \)
To find when \(K_c\) and \(K_p\) are numerically equal, the equation simplifies if \((RT)^{\Delta n} = 1\). This condition determines the scenarios where the constants are equal. Understanding this relationship helps predict how changes in conditions affect the state of equilibrium in a system.
Ideal Gas Constant
The Ideal Gas Constant, denoted as \( R \), plays a crucial role in converting between \( K_c \) and \( K_p \). Its standard value used in these calculations is 0.0821 L atm K鈦宦� mol鈦宦�. This constant appears in the ideal gas equation \(PV = nRT\), linking pressure \(P\), volume \(V\), moles of gas \(n\), and temperature \(T\).

Including \( R \) in the relationship between \( K_c \) and \( K_p \):
  • The term \((RT)^{\Delta n}\) highlights the impact temperature and change in moles have on the equilibrium constants.
  • In conditions where \((RT)^{\Delta n} = 1\), \( R \) helps determine at what temperature \(T\) the reaction maintains equilibrium constants as equal.
Understanding \( R \) ensures accurate application in calculating gas behaviors, especially in relation to chemical equilibria involving gases.
Change in Moles of Gas (螖n)
The change in moles of gas, expressed as \( \Delta n \), is the difference between the number of moles of gaseous products and gaseous reactants in a balanced chemical equation. It significantly affects the relationship between \( K_c \) and \( K_p \).

When \( \Delta n = 0 \), it means that there is no net change in gas moles during the reaction. This occurs when the number of gaseous molecules remains the same as reactants become products. In such a case:
  • \((RT)^{\Delta n} = (RT)^0 = 1\)
  • \(K_c\) equals \(K_p\) regardless of the temperature \(T\) or the value of \( R \).
If \( \Delta n \) is not zero, temperature adjustments can be made to attain this equality, specifically by selecting \( T \) such that \((RT)^{\Delta n} = 1\).

Recognizing the role of \( \Delta n \) aids in predicting how changes in temperature and pressure influence equilibria in gas-phase reactions.

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Most popular questions from this chapter

Does adding reactants to a system at equilibrium increase the value of the equilibrium constant?

A 100 mL reaction vessel initially contains \(2.60 \times 10^{-2} \mathrm{mol} \mathrm{NO}\) and \(1.30 \times 10^{-2} \mathrm{mol} \mathrm{H}_{2}, \mathrm{At}\) cquilibrium, the concentration of NO in the vesscl is \(0.161 M .\) What is the value of \(K_{\mathrm{c}}\) for the following reaction? $$ 2 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g) $$

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Reactions between carboxylic acids and alcohols to produce esters typically do not go to completion. Ethyl acetate, a compound used industrially to decaffeinate coffee and tea, is made in the following reaction for which \(K_{c}=3.87\) at \(75^{\circ} \mathrm{C}\) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(\ell)+\mathrm{CH}_{3} \operatorname{cooH}(\ell) \rightleftharpoons\) $$ \mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}(e)+\mathrm{H}_{2} \mathrm{O}(\ell) $$ Is a mixture of \(125 \mathrm{g}\) of ethanol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right), 125 \mathrm{g}\) of acctic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right), 125 \mathrm{g}\) of cthyl acctate \(\left(\mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}\right),\) and \(125 \mathrm{g}\) of watcr in a \(5 \mathrm{L}\) reactor at equilibrium? b. If not, in which direction will the reaction shift to reach equilibrium?
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