Problem 134 - A Complete Resource Book in Chemistry for JEE Main

Question

For the 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 \(\mathrm{K}_{n}\) is \(1.44 \times 10^{-5}\) when the partial pressures are measured in atmosphere. The value of \(\mathrm{K}_{\mathrm{c}}\) with concentration in \(\mathrm{mol} \mathrm{L}^{-1}\) is (a) \(\frac{1.44 \times 10^{-5}}{(8.314 \times 773)^{-2}}\) (b) \(\frac{1.44 \times 10^{-5}}{(0.082 \times 500)^{-2}}\) (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 expression for \(\mathrm{K}_c\) is given by option (c) or (d).

Step 1: Understanding the Reaction and Given Data

The reaction is \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\). The given \(\mathrm{K}_n\) value is \(1.44 \times 10^{-5}\) at \(500^{\circ} \mathrm{C}\), with pressures in atmospheres. \(\mathrm{K}_c\) is the equilibrium constant expressed in molarity (\(\mathrm{mol} \mathrm{L}^{-1}\)). We need to find the correct option for \(\mathrm{K}_c\).

Step 2: Relationship Between \(\mathrm{K}_n\) and \(\mathrm{K}_c\)

The relationship between \(\mathrm{K}_n\) and \(\mathrm{K}_c\) is given by the equation \(\mathrm{K}_c = \mathrm{K}_n \times (RT)^{-\Delta n}\), where \(R\) is the ideal gas constant, \(T\) is the temperature in Kelvin, and \(\Delta n\) is the change in moles of gas from reactants to products.

Step 3: Calculating \(\Delta n\)

For the reaction \(\mathrm{N}_{2}(\mathrm{~g}) + 3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\), \(\Delta n = (2) - (1+3) = -2\).

Step 4: Determine the Temperature in Kelvin

Convert the temperature from Celsius to Kelvin: \(T = 500^{\circ} \mathrm{C} + 273 = 773 \mathrm{~K}\).

Step 5: Apply the Formula for \(\mathrm{K}_c\)

Substitute the given values into the formula: \(\mathrm{K}_c = \mathrm{K}_n \times (RT)^{-\Delta n}\). This becomes \(\mathrm{K}_c = 1.44 \times 10^{-5} \times (0.0821 \times 773)^{2}\) because \(-\Delta n = 2\).

Step 6: Simplify the Expression

Calculate \((0.0821 \times 773)\) and square the result. The squared value is used as a divisor according to option (c). Thus, \(\mathrm{K}_c = \frac{1.44 \times 10^{-5}}{(0.0821 \times 773)^2}\), matching options (c) and (d).

Key Concepts

Reaction Thermodynamics

Thermodynamics helps us understand how and why reactions occur. It involves studying the changes in energy and the directionality of chemical processes. In thermodynamics, we are interested in properties such as enthalpy (heat content), entropy (degree of disorder), and Gibbs free energy (useful for predicting spontaneity of reactions).
For any given reaction, these properties tell us if it can proceed by itself or if it needs an external source of energy.

• Enthalpy (\( \Delta H \)): Released or absorbed heat during a reaction. Negatively valued enthalpy indicates an exothermic reaction, releasing heat.
• Entropy (\( \Delta S \)): Measures the spread of energy in a system, with positive \( \Delta S \) suggesting an increase in disorder or energy dispersion.
• Gibbs Free Energy (\( \Delta G \)): Combines enthalpy and entropy. If \( \Delta G \) is negative, the reaction is spontaneous.

In this reaction where nitrogen and hydrogen form ammonia, we are interested in the equilibrium constant (\( K \)), which is reflected in \( K_n \) and \( K_c \) based on pressure and concentration, respectively.
Understanding thermodynamics helps determine the energy landscape over which these reactions occur.

Chemical Equilibrium

Chemical equilibrium signifies a state where the forward and reverse reactions occur at the same rate. In other words, the concentration of reactants and products remains constant over time.
For the ammonia synthesis reaction \( \mathrm{N}_{2}(\mathrm{g}) + 3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{g}) \), equilibrium doesn't mean that the amounts of nitrogen, hydrogen, and ammonia are equal. Instead, it means their rates of conversion to other forms are balanced.

• The equilibrium constant, \( K \), quantifies the relative quantities of reactants and products at equilibrium.
• \( K_n \) is used for when the reaction uses partial pressures.
• \( K_c \) is used when concentrations are measured in molarity.

At equilibrium, adjusting variables like temperature or pressure tends to shift the balance as explained by Le Chatelier’s principle. For example, increasing temperature generally favors the endothermic reaction direction if any. Calculating \( K_c \) from \( K_n \) requires understanding the difference in the units and how pressure relates to concentration, often involving the Ideal Gas Law.

Ideal Gas Law

The Ideal Gas Law is a fundamental principle that connects pressure (\( P \)), volume (\( V \)), temperature (\( T \)), and the amount of substance (\( n \)) as \( PV = nRT \), where \( R \) is the ideal gas constant. This relationship is especially useful when dealing with gases, such as nitrogen and hydrogen in our example reaction.
To switch from partial pressures to molar concentrations (needed to compute \( K_c \)), we use this law. At a constant temperature and volume, this law allows us to say that pressure is directly related to concentration. This relationship can be expressed as:

• Concentration (\( C \)) in moles per liter results from dividing the pressure by \( RT \).

Using this principle, we convert between \( K_n \) and \( K_c \) by integrating temperature and volume details. This is crucial when calculating equilibrium states in reactions involving gases, as it allows us to derive concentrations from measurable pressures. For the ammonia synthesis, the conversion uses the ideal gas constant, temperature in Kelvin, and the stoichiometric changes in molar quantities.