Problem 147 - A Complete Resource Book in Chemistry for JEE Main

Question

\(\frac{\mathrm{K}_{\mathrm{p}}}{\mathrm{K}_{\mathrm{C}}}\) for the reaction \(\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})\) will be (a) \(\sqrt{\mathrm{RT}}\) (b) \(\mathrm{RT}\) (c) \(\frac{1}{\sqrt{\mathrm{RT}}}\) (d) 1

Step-by-Step Solution

Answer

The correct choice is (c) \(\frac{1}{\sqrt{RT}}\).

Step 1: Identify Known Variables and Basic Equation

The equilibrium constants for reactions involving gases can be related using the relation: \[ K_p = K_c(RT)^n \], where \(n\) is the change in number of moles of gases (moles of gaseous products minus moles of gaseous reactants) and \(R\) is the ideal gas constant. Given: \( CO(g) + \frac{1}{2}O_2(g) \rightleftharpoons CO_2(g) \).

Step 2: Calculate Change in Moles (n)

Determine the change in the number of moles of gases during the reaction. Here, reactants include 1 mole of CO and 0.5 moles of O2 (total 1.5 moles), while the product is 1 mole of CO2. Thus, \(n = 1 - 1.5 = -0.5\).

Step 3: Substitute into the Kp and Kc Relationship

Use \[ K_p = K_c (RT)^n \] to calculate the ratio \( \frac{K_p}{K_c} \). Given \( n = -0.5 \), this becomes \[ K_p = K_c (RT)^{-0.5} \]. Therefore, \[ \frac{K_p}{K_c} = (RT)^{-0.5} = \frac{1}{\sqrt{RT}} \].

Step 4: Interpret and Conclude

The calculated result \( \frac{K_p}{K_c} = \frac{1}{\sqrt{RT}} \) matches option (c) in the list of possible answers.

Key Concepts

Kp and Kc relation

In chemical reactions involving gases, equilibrium constants can be expressed in terms of partial pressures, known as \(K_p\), or concentrations, known as \(K_c\). Understanding the relationship between these two constants is key for solving equilibrium problems. The general formula that connects \(K_p\) and \(K_c\) is: \[ K_p = K_c (RT)^n \] where:

• \(K_p\) refers to the equilibrium constant in terms of partial pressures.
• \(K_c\) refers to the equilibrium constant in terms of concentrations.
• \(R\) is the ideal gas constant.
• \(T\) is the temperature in Kelvin.
• \(n\) is the change in moles of gas, calculated as moles of gaseous products minus moles of gaseous reactants.

This equation helps in converting between these constants when solving chemical equilibrium problems, allowing you to understand how the equilibrium of a gas reaction changes with different conditions.
Exploring this relationship can provide insight into how gases behave under equilibrium, especially when reacting species are all gases.

Change in moles calculation

The change in moles \(n\) is a critical aspect when dealing with the relationship between \(K_p\) and \(K_c\). Calculating \(n\) involves determining the difference in the number of moles of gases produced and consumed in a reaction. In the given problem, the reaction is \(CO(g) + \frac{1}{2}O_2(g) \rightleftharpoons CO_2(g)\).

• Reactants: 1 mole of CO and 0.5 moles of \(O_2\), totaling 1.5 moles of gaseous reactants.
• Products: 1 mole of \(CO_2\).
• Thus, \(n = 1 - 1.5 = -0.5\).

The value of \(n\) is negative in this case, indicating a decrease in the total number of moles of gas as the reaction proceeds from reactants to products.
This calculation is important because it directly affects the relation between \(K_p\) and \(K_c\), transforming their relationship through the term \((RT)^n\). Without understanding the calculation of \(n\), predicting the behavior of the equilibrium constants could be impossible.

Ideal gas constant

The ideal gas constant \(R\) plays a significant role in the formula relating \(K_p\) and \(K_c\). The constant \(R\) is a fundamental value in chemistry that links various gas-related equations and calculations. Here we express it as part of the equation \(K_p = K_c(RT)^n\), affecting how equilibrium constants are converted.
\(R\) can take various values such as 0.0821 L·atm/mol·K or 8.314 J/mol·K, depending on the units used in the problem.

• \(R\) helps adjust the equilibrium constant formula to account for temperature effects since the terms are raised to the power of \(n\).
• The choice of \(R\) impacts the numerical value of equilibrium calculations, linking pressure and concentration scales.

By understanding and correctly using \(R\), students can ensure their equilibrium calculations are accurate, no matter the conditions or the specific setup of their chemical equations. Proper use of the ideal gas constant is essential for solving chemical equilibrium problems involving gases.