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 \(\mathrm{K}_{\mathrm{p}}\) 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 answer is (c) \(\frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}}\).

Step 1: Identify Reaction and Related Constants

The reaction given is \(\mathrm{N}_{2}(\mathrm{~g}) + 3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\). We are given \(\mathrm{K}_{\mathrm{p}} = 1.44 \times 10^{-5}\) at \(500^{\circ} \mathrm{C}\), and we need to find \(\mathrm{K}_{\mathrm{c}}\).

Step 2: Determine the Correlation Between Kp and Kc

Use the formula \( K_{\mathrm{p}} = K_{\mathrm{c}} \times (R \times T)^{\Delta n} \), where \(R\) is the gas constant (0.0821 L atm K⁻¹ mol⁻¹ or 8.314 J K⁻¹ mol⁻¹), \(T\) is the temperature in Kelvin, and \(\Delta n\) is the change in moles of gas.

Step 3: Calculation of Δn

\(\Delta n\) is calculated by determining the moles of gaseous products minus the moles of gaseous reactants: \(2 - (1 + 3) = -2\). Thus, \(\Delta n = -2\).

Step 4: Convert Temperature to Kelvin

The temperature in Celsius is given as \(500^{\circ} \mathrm{C}\). Convert it to Kelvin by adding 273: \(500 + 273 = 773 \mathrm{~K}\).

Step 5: Substitute Values into Kp-Kc Equation

Substitute the known values into the equation: \[ 1.44 \times 10^{-5} = K_{\mathrm{c}} \times (0.0821 \times 773)^{-2} \]

Step 6: Solve for Kc

Rearrange the equation to solve for \(K_{\mathrm{c}}\): \[ K_{\mathrm{c}} = \frac{1.44 \times 10^{-5}}{(0.0821 \times 773)^{-2}} \]

Step 7: Verify Choices Against Calculation

Check against the options provided:- Option (c) and (d) suggests the same expression, and matches the calculated expression for \(K_{\mathrm{c}}\). Therefore, both options are correct according to the given problem.

Key Concepts

Equilibrium Constants

The concept of equilibrium constants is at the center of understanding chemical equilibria. In the reaction \( \mathrm{N}_{2}(\mathrm{~g}) + 3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \), we use equilibrium constants to quantify the position of the equilibrium.

• \( \mathrm{K}_{\mathrm{c}} \): Represents the equilibrium constant in terms of concentration with units of \( \text{mol} \cdot \text{L}^{-1} \).
• \( \mathrm{K}_{\mathrm{p}} \): Represents the equilibrium constant in terms of partial pressures with units of atmosphere (atm).In equations, \( K_{\mathrm{c}} \) and \( K_{\mathrm{p}} \) are related by the expression: \[K_{\mathrm{p}} = K_{\mathrm{c}} \times (R \times T)^{\Delta n}\]where \( R \) is the ideal gas constant, \( T \) is temperature, and \( \Delta n \) is the change in mols of gas.Understanding equilibrium constants helps us predict which direction a reaction needs to move to achieve equilibrium.

Reaction Quotient

The reaction quotient, noted as \( Q \), allows us to understand if a reaction is at equilibrium or not, and if not, in which direction it needs to proceed to reach it.

• If \( Q = K \): The system is at equilibrium.
• If \( Q < K \): More reactants will react to form products.
• If \( Q > K \): More products will revert to form reactants.For the reaction \( \mathrm{N}_{2}(\mathrm{~g}) + 3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \), we can calculate \( Q \) the same way as \( K \), using initial concentrations or partial pressures:\[Q = \frac{{P_{\text{\textrm{NH}}_{3}}}^2}{P_{\text{\textrm{N}}_{2}} \times P_{\text{\textrm{H}}_{2}}^3}\]The comparison of \( Q \) to \( K \) is crucial in determining the state of the reaction and predicting how it will proceed.

Gas Equilibrium

Gas equilibrium involves reactions where gaseous reactants and products are involved, like in our given reaction. Such equilibria depend heavily on pressure and volume due to the properties of gases.

• The Equilibrium constant \( K_{\mathrm{p}} \) is specifically used for gas equilibrium when pressure is measured in atmospheres.
• The change in the number of moles\( (\Delta n) \) affects how pressure influences equilibrium position.Key to gas equilibrium is the computation of \( \Delta n \), calculated as: \[\Delta n = \text{moles of products} - \text{moles of reactants}\]For our reaction, \( \Delta n \) is \(-2\) meaning a decrease in total moles when the reaction proceeds to the right. Thus, reducing pressure favors product formation when \( \Delta n \) is negative.

Thermodynamics

Thermodynamics questions about how energy changes during chemical reactions and how these changes affect equilibrium.

• The Gibbs Free Energy change, \( \Delta G \), indicates the spontaneity of a reaction.
• At equilibrium, \( \Delta G = 0 \), meaning there is no net change in energy.The relation of \( \Delta G \) to equilibrium constant \( K \) is: \[\Delta G = -RT \ln{K}\]where a negative \( \Delta G \) suggests a reaction proceeds spontaneously in that direction. A smaller or larger \( K \) value can hint at whether the forward or reverse reaction is favored under constant thermodynamic conditions.In sum, thermodynamics intertwines with equilibrium to help us determine how energy landscapes drive chemical behavior.