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Chapter 7: Problem 79

The value of \(K_{p}\) for the reaction, \(2 \mathrm{SO}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3}\) at 700 is \(1.3 \times 10^{-3} \mathrm{~atm}^{-1} .\) The value of \(K_{c}\) at same temperature will be (a) \(1.4 \times 10^{-2}\) (b) \(7.4 \times 10^{-2}\) (c) \(5.2 \times 10^{-2}\) (d) \(3.1 \times 10^{-2}\)

Short Answer

Expert verified
The value of \(K_c\) is approximately \(7.4 \times 10^{-2}\), so the correct option is (b).

Step by step solution

01

Understanding the Relationship Between Kc and Kp

The relationship between the equilibrium constants \(K_c\) and \(K_p\) is expressed using the equation:\[ K_p = K_c (RT)^{\Delta n} \]where \( R \) is the ideal gas constant (0.0821 L atm/mol K), \( T \) is the temperature in Kelvin, and \( \Delta n \) is the change in moles of gas (products - reactants).
02

Calculate \(\Delta n\)

For the reaction \(2 \mathrm{SO}_2 + \mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3\), the change in moles \( \Delta n \) is calculated as:\[ \Delta n = (2) - (2 + 1) = 2 - 3 = -1 \]
03

Rearrange the Equation

Since we have \(\Delta n = -1\), we rearrange the equation to solve for \(K_c\):\[ K_c = \frac{K_p}{(RT)^{-1}} \]This simplifies to:\[ K_c = K_p \cdot (RT) \]
04

Substitute Values and Calculate \(K_c\)

Given \(K_p = 1.3 \times 10^{-3} \ \text{atm}^{-1}\), the ideal gas constant \(R = 0.0821\ \text{L atm/mol K}\), and temperature \(T = 700\ K\), substitute the values into the equation:\[ K_c = 1.3 \times 10^{-3} \times (0.0821\times 700) \]Calculate the value:\[ K_c = 1.3 \times 10^{-3} \times 57.47 = 7.4711 \times 10^{-2}\]
05

Choose the Correct Option

Compare the calculated \( K_c \) value of \( 7.4711 \times 10^{-2} \) to the given options. The closest match is option (b) \( 7.4 \times 10^{-2} \).

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Key Concepts

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

Kp and Kc relationship
Equilibrium constants, specifically \( K_p \) and \( K_c \), are two important parameters in chemical reactions at equilibrium. They help quantify the extent of a reaction at a certain temperature. These constants, although both related to equilibrium, cater to different aspects. While \( K_c \) is the equilibrium constant expressed in terms of concentrations (typically in mol/L), \( K_p \) is expressed in terms of partial pressures (typically in atm).

The relationship between them is beautifully conveyed by the equation:
  • \( K_p = K_c (RT)^{\Delta n} \)
Where:
  • \( R \) is the ideal gas constant
  • \( T \) is the temperature in Kelvin
  • \( \Delta n \) denotes the change in moles between reactants and products
This equation highlights that the relationship between \( K_p \) and \( K_c \) is temperature-dependent and considers the change in moles during the reaction.
ideal gas constant
The ideal gas constant, denoted as \( R \), is a fundamental constant that appears in the ideal gas law, PV = nRT, where it connects pressure, volume, and temperature for an ideal gas. Its value is 0.0821 L atm/mol K.

This constant is crucial for relating the equilibrium constants \( K_c \) and \( K_p \), helping translate between concentrations and partial pressures through the equation:
  • \( K_p = K_c (RT)^{\Delta n} \)
The constant \( R \) allows for conversions needed in chemical equations involving gases, keeping measurements consistent when dealing with varying reaction conditions. Hence, when performing calculations for equilibrium constants, using the correct value and unit for \( R \) is essential to ensure accuracy.
change in moles (螖n)
The term \( \Delta n \) refers to the change in the number of moles of gas as a reaction progresses from reactants to products. It is an essential component in calculating the relationship between \( K_c \) and \( K_p \).

To calculate \( \Delta n \), subtract the sum of the stoichiometric coefficients of the gaseous reactants from that of the gaseous products:
For example, in the reaction:
  • \( 2 \text{SO}_2 + \text{O}_2 \rightleftharpoons 2 \text{SO}_3 \)
Then \( \Delta n \) can be calculated as:
  • \( \Delta n = 2 - (2 + 1) = -1 \)
Understanding \( \Delta n \) helps predict how pressure changes affect the equilibrium position and allows for accurate calculations when dealing with reactions involving gases. A negative \( \Delta n \) indicates a decrease in moles of gas, showing the reaction shifts towards fewer gaseous molecules.
temperature in Kelvin
When working with reactions and equilibrium constants, it is essential to use the temperature in Kelvin. This is because Kelvin is the absolute temperature scale which starts at absolute zero, making it crucial for thermodynamic calculations. Using Kelvin eliminates negative numbers that could occur with the Celsius scale and aligns with the requirements of the ideal gas law and related equations.

In the context of the relationship between \( K_p \) and \( K_c \), the temperature \( T \) in Kelvin is a pivotal part of the equation:
  • \( K_p = K_c (RT)^{\Delta n} \)
In our example, where the temperature is provided as 700 K, we utilize this value directly in our calculations to derive \( K_c \) from \( K_p \), ensuring precision and consistency across thermodynamic evaluations.

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Most popular questions from this chapter

\(1.25\) moles of NOCl were placed in a \(2.50 \mathrm{~L}\) reaction chamberat \(427^{\circ} \mathrm{C}\). After equilibrium was reached, \(1.10\) moles of \(\mathrm{NOCl}\) remained. Calculate the equilibrium constant \(K_{c}\) for the reaction, \(2 \mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (a) \(1.6 \times 10^{-3}\) (b) \(5.6 \times 10^{-4}\) (c) \(2.6 \times 10^{-3}\) (d) \(4.6 \times 10^{-4}\)

The equilibrium constant of mutarotation of \(\alpha\)-D-glucose to \(\beta\)-D-glucose is \(1.8\). What per cent of the \(\alpha\)-form remains under equilibrium? (a) \(35.7\) (b) \(64.3\) (c) \(55.6\) (d) \(44.4\)

Phosphorous pentachloride dissociates as follows, in a closed reaction vessel \(\mathrm{PCI}_{5}(\mathrm{~g}) \longrightarrow \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) If total pressure at equilibrium of the reaction mixture is \(\mathrm{P}\) and degree of dissociation of \(\mathrm{PCl}_{5}\) is \(\mathrm{x}\), the partial pressure of \(\mathrm{PCl}_{3}\) will be (a) \(\left(\frac{x}{(x+1)}\right) \mathrm{P}\) (b) \(\left(\frac{2 x}{(x-1)}\right) \mathrm{P}\) (c) \(\left(\frac{x}{(x-1)}\right) \mathrm{P}\) (d) \(\left(\frac{x}{(1-x)}\right) \mathrm{P}\)

When solid \(\mathrm{NaNO}_{3}\) is heated in a closed vessel, \(\mathrm{O}_{2}\) is liberated and solid \(\mathrm{NaNO}_{2}\) is left behind. At equilibrium (a) Addition of \(\mathrm{NaNO}_{2}\) favours reverse reaction. (b) Addition of \(\mathrm{NaNO}_{3}\) favours forward reaction. (c) Increasing the pressure favours reverse reaction. (d) Increasing the temperature favours forward reaction.

At certain temperature compound \(\mathrm{AB}_{2}(\mathrm{~g})\) dissociates according to the reaction \(2 \mathrm{AB}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{g})+\mathrm{B}_{2}(\mathrm{~g})\) The value of \(K_{p}\) in terms of degree of dissociation ' \(\alpha\) ' and total pressure ' \(\mathrm{P}\) ' is (a) \(\mathrm{P} \frac{\alpha^{3}}{2}\) (b) \(\mathrm{P} \frac{\alpha^{2}}{3}\) (c) \(\mathrm{P} \frac{\alpha^{3}}{3}\) (d) \(\mathrm{P} \frac{\alpha^{2}}{2}\)
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