Problem 27 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

For the reversible reaction, \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) at \(500^{\circ} \mathrm{C}\), the value of \(K_{p}\) is \(1.44 \times 10^{-5}\) when partial pressure is measured in atmospheres. The corresponding value of \(K_{c}\), with concentration in mole litre \(^{-1}\), is (a) \(\frac{1.44 \times 10^{-5}}{(0.082 \times 500)^{-2}}\) (b) \(\frac{1.44 \times 10^{-5}}{(8.314 \times 773)^{-2}}\) (c) \(\frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{2}}\) (d) \(\frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}}\)

Step-by-Step Solution

Answer

The correct answer is (d).

Step 1: Understand the Relationship between Kc and Kp

To convert between the equilibrium constant based on partial pressures \(K_p\) and the equilibrium constant based on concentration \(K_c\), we use the equation: \[ K_p = K_c (RT)^{\Delta n} \] where \(\Delta n\) is the change in moles of gas, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin.

Step 2: Determine \(\Delta n\) and Temperature Conversion

Here, \(\Delta n\) is calculated as moles of gaseous products minus moles of gaseous reactants. For the reaction, \(\Delta n = 2 - (1+3) = -2\). The temperature given is \(500^{\circ}C\), which we convert to Kelvin as follows: \[ T(K) = 500 + 273 = 773 \, K \]

Step 3: Plug Values into the Equation

Substitute \(K_p = 1.44 \times 10^{-5}\), \(\Delta n = -2\), \(R = 0.082 \, L\,atm\,K^{-1}\,mol^{-1}\), and \(T = 773 \, K\) into the relationship: \[ K_c = \frac{K_p}{(RT)^{\Delta n}} = \frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}} \]

Step 4: Identify the Correct Option

The expression from Step 3, \(\frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}}\), matches option (d).

Key Concepts

Reversible Reactions

In chemistry, a reversible reaction is a type of reaction where the products can convert back into the original reactants. These reactions are represented by the double arrow symbol \[ \rightleftharpoons \]This symbol indicates that both forward and reverse reactions are occurring simultaneously. A classic example is the synthesis of ammonia from nitrogen and hydrogen gases:\[\mathrm{N}_{2}( ext{g}) + 3\mathrm{H}_{2}( ext{g}) \rightleftharpoons 2\mathrm{NH}_{3}( ext{g})\]The conditions such as pressure, temperature, and concentration of reactants and products influence whether the reaction proceeds predominantly in one direction or remains in equilibrium. This equilibrium occurs when the rate of the forward reaction equals the rate of the backward reaction, resulting in constant concentrations of reactants and products. Understanding these dynamics is essential for manipulating conditions in industrial processes to maximize product yield.

Kc and Kp Relationship

Equilibrium constants provide a quantitative measure of the position of equilibrium in a chemical reaction. Two commonly used forms of equilibrium constants are:- **\(K_c\):** The equilibrium constant when concentrations are expressed in moles per liter.- **\(K_p\):** The equilibrium constant when using partial pressures of gases, typically in atmospheres.The relationship between \(K_c\) and \(K_p\) is given by the equation:\[K_p = K_c (RT)^{\Delta n}\]Here:- \(R\) is the universal gas constant, typically 0.0821 L atm K⁻¹ mol⁻¹ or 8.314 J K⁻¹ mol⁻¹ (depending on the units used).- \(T\) is the temperature in Kelvin.- \(\Delta n\) is the difference in moles of gaseous products and reactants.This formula reflects the dependency of the equilibrium constant form on temperature and pressure, emphasizing the need to understand the context in which each is applied. Changes in pressure and temperature can influence \(\Delta n\), thus altering \(K_p\) and \(K_c\).

Gas Constant and Temperature Conversions

To correctly apply the formulas for chemical equilibrium constants, understanding the universal gas constant \(R\) and proper temperature conversions is crucial. **Universal Gas Constant \(R\):**- \(R\) has different values and units depending on the context: - 0.0821 L atm K⁻¹ mol⁻¹, used when dealing with gases in atmospheres. - 8.314 J K⁻¹ mol⁻¹, applicable in energy-related calculations.**Temperature Conversions:**- Chemical calculations often require the temperature to be in Kelvin for consistency.- Convert Celsius to Kelvin using: \[ T(K) = T(°C) + 273.15 \]In our example exercise, knowing the temperature at 500°C converts to 773 K is critical for deciding the correct unit for \(R\). Applying consistent units ensures the accuracy of equilibrium calculations like those needed for converting between \(K_c\) and \(K_p\). Understanding these basics can significantly ease solving typical textbook exercises in equilibrium chemistry.