Problem 194 - A Complete Resource Book in Chemistry for JEE Main

Question

For the reaction, \(2 \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\) \(\left(\mathrm{K}_{c}=1.8 \times 10^{-6}\right.\) at \(\left.184^{\circ} \mathrm{C}\right)\) \((\mathrm{R}=0.0831 \mathrm{~kJ} /(\mathrm{mol} \mathrm{K}))\) when \(\mathrm{K}_{\mathrm{p}}\) and \(\mathrm{K}_{c}\) are compared at \(184^{\circ} \mathrm{C}\) it is found that (a) \(\mathrm{K}_{\mathrm{p}}\) is greater than \(\mathrm{K}_{\mathrm{c}}\) (b) \(\mathrm{K}_{\mathrm{p}}\) is less than \(\mathrm{K}_{\mathrm{c}}\) (c) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}\) (d) whether \(\mathrm{K}_{\mathrm{p}}\) is greater than, less than or equal to \(\mathrm{K}_{c}\) depends upon the total gas pressure

Step-by-Step Solution

Answer

Kp is greater than Kc; option (a) is correct.

Step 1: Understand the Relationship Between Kp and Kc

Before comparing \( K_p \) and \( K_c \), recall the relationship between them: \( K_p = K_c (RT)^{\Delta n} \).Where \( R \) is the ideal gas constant, \( T \) is the temperature in Kelvin, and \( \Delta n \) is the change in the number of moles of gas (products - reactants).

Step 2: Calculate Delta n

Identify \( \Delta n \) for the reaction. For \( 2 \text{NO}_2(g) \rightleftharpoons 2 \text{NO}(g) + \text{O}_2(g) \), there are 3 moles of products (2 \text{NO} + 1 \text{O}_2) and 2 moles of reactants (2 \text{NO}_2). Therefore, \( \Delta n = 3 - 2 = 1 \).

Step 3: Convert Temperature to Kelvin

To apply the formula, convert the temperature from Celsius to Kelvin: \( T = 184 + 273.15 = 457.15 \text{ K} \).

Step 4: Calculate Kp

Using the formula \( K_p = K_c (RT)^{\Delta n} \), substitute \( K_c = 1.8 \times 10^{-6} \), \( R = 0.0831 \), \( T = 457.15 \text{ K} \), and \( \Delta n = 1 \):\[ K_p = 1.8 \times 10^{-6} \times (0.0831 \times 457.15)^1 \].This simplifies to \[ K_p = 1.8 \times 10^{-6} \times 37.967815 \approx 6.834 \times 10^{-5} \].

Step 5: Compare Kp and Kc

Now, compare the values: \( K_p = 6.834 \times 10^{-5} \) and \( K_c = 1.8 \times 10^{-6} \). Since \( K_p > K_c \), option (a) is correct.

Key Concepts

Kp and Kc relationship

In chemical equilibrium, understanding the connection between pressure equilibrium constant, \( K_p \), and concentration equilibrium constant, \( K_c \), is crucial.
These constants give us insight into how reactions behave under changes in temperature and pressure. The mathematical relation between these two is expressed as:

• \( K_p = K_c (RT)^{\Delta n} \)

Here, \( R \) signifies the ideal gas constant, \( T \) is the temperature (which must be in Kelvin for this equation), and \( \Delta n \) is the change in moles of gas.
This relationship helps to predict how gases' equilibrium is influenced when switching between measurements in concentration and pressure. A key takeaway is that whenever \( \Delta n eq 0 \), \( K_p \) will generally differ from \( K_c \). This highlights the importance of knowing the moles in both products and reactants.

Delta n calculation

The term \( \Delta n \) denotes the change in the number of moles of gases and plays a pivotal role in the \( K_p \) and \( K_c \) relationship.
For any gas-phase reaction, understanding \( \Delta n \) is essential for calculations involving equilibrium constants.
To calculate \( \Delta n \), follow these steps:

• Identify the total number of moles of gaseous products.
• Identify the total number of moles of gaseous reactants.
• Subtract the moles of reactants from the moles of products to get \( \Delta n \).

For example, consider the reaction \( 2 \text{NO}_2(g) \rightleftharpoons 2 \text{NO}(g) + \text{O}_2(g) \).
In this case, there are 3 moles of products and 2 moles of reactants, so \( \Delta n = 3 - 2 = 1 \).
Understanding and accurately calculating \( \Delta n \) is vital because it directly affects the computation of \( K_p \) from \( K_c \).

Temperature conversion to Kelvin

In the realm of chemical equations, particularly those involving gases, temperatures must be converted to Kelvin.
Why? Because Kelvin is the absolute temperature scale, ensuring that all related calculations are consistent with the laws of thermodynamics.
The conversion from Celsius to Kelvin is straightforward:

• Add 273.15 to the Celsius temperature.

For instance, a temperature of \( 184^{\circ}\text{C} \) becomes:

• \( 184 + 273.15 = 457.15 \text{ K} \)

This conversion is imperative because many equations, including the one relating \( K_p \) and \( K_c \), require temperatures in Kelvin. Using Kelvin ensures that the ideal gas constant, \( R \), applies correctly, giving accurate results for reactions and their equilibrium constants.