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 \(\mathrm{K}_{\mathrm{n}}\) is \(1.44 \times 10^{-5}\) when the partial pressures are measured in atmosphere. The value of \(\mathrm{K}_{\text {s }}^{\text {p }}\) 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).

Step 1: Identify the Relationship

We need to relate the equilibrium constant in terms of pressure, \(K_p\), to the equilibrium constant in terms of concentration, \(K_c\). The relationship between \(K_p\) and \(K_c\) is given by the equation: \[ K_p = K_c (RT)^{\Delta n} \] where \(\Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants}.\) Here, for the reaction, \(\Delta n = 2 - (1 + 3) = -2\).

Step 2: Calculate the Components

First, determine the \(\Delta n\), which we have found to be \(-2\). Next, identify the temperature \(T\) in Kelvin: \(T = 500^{\circ} \text{C} + 273 = 773\, K\). \(R\) is the ideal gas constant, which is 0.0821 L atm K^{-1} mol^{-1} when dealing with atmospheres.

Step 3: Express K_s in Terms of K_n

From the equation \( K_p = K_c (RT)^{\Delta n} \), rearrange to get \( K_c = \frac{K_p}{(RT)^{\Delta n}} \). Substitute \(K_n\) for \(K_p\), resulting in \( K_c = \frac{1.44 \times 10^{-5}}{(0.0821 \times 773)^{-(-2)}} \).

Step 4: Calculate the Expression

Compute the value of the expression: \( (0.0821 \times 773)^2 \). This product equals about 5,054.041. Substitute this value back to find \( K_c \): \( K_c = \frac{1.44 \times 10^{-5}}{5054.041} \), resulting in \( K_c \approx 2.85 \times 10^{-9} \).

Step 5: Match the Correct Answer

Looking at the provided options, note that each involves a form of the expression \( \frac{1.44 \times 10^{-5}}{(0.082 \times T)^2} \). Option (c) matches this expression as it shows the conversion using the ideal gas constant and temperature: \(\frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}}\). Given the calculations, the answer is option (c).

Key Concepts

Chemical Equilibrium

During a chemical reaction, we often reach a state known as chemical equilibrium. This is when the forward and reverse reactions occur at the same rate. Imagine a busy street—cars move in both directions consistently; no lane gets more crowded over time. This balanced flow is similar to a chemical system at equilibrium.

For any chemical reaction, like the \[ \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}), \]we can describe the equilibrium using an equilibrium constant. When measuring in terms of pressure, we call this \( K_p \).

• Equilibrium constants help predict product or reactant dominance.
• A small \( K_p \) indicates more reactants than products.

Ideal Gas Law

The ideal gas law relates a gas's pressure, volume, temperature, and number of moles. It simplifies complexities we find in real gases. The relation is captured by the formula \( PV = nRT \).

Here's what each part means:

• \( P \) is the pressure, usually in atmospheres, measured by how much the gas particles bump into the walls of their container.
• \( V \) is the volume, the space the gas occupies.
• \( n \) stands for the number of moles, a count of particles.
• \( R \), the ideal gas constant, has a value of \( 0.0821 \) L atm K\( ^{-1} \) mol\( ^{-1} \).
• \( T \) is the temperature in Kelvin.

This law is crucial when converting between pressure-based and concentration-based equilibrium constants, using \( RT \).

Reaction Kinetics

Understanding reaction kinetics is like getting to know the speed limits and traffic flow rules of our chemistry roads. It explores how fast or slow a reaction occurs. The speed is affected by several factors.

• Concentration: More particles lead to more collisions and faster reactions.
• Temperature: Higher temperatures usually mean more energetic collisions.
• Pressure: In the case of gases, increased pressure pushes particles closer together, quickening reactions.
• Catalysts: These are substances that speed up a reaction without changing themselves.

Ultimately, kinetics and equilibrium differ in focus—kinetics concerns itself with how quickly equilibrium is achieved.

Le Chatelier's Principle

Le Chatelier's principle acts like the GPS for a chemical equation at equilibrium. It tells us how a system shifts to new conditions. If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change.

• Adding reactants or products can shift the equilibrium to favor the opposite side.
• Changes in temperature alter the balance, depending on whether the reaction is endothermic or exothermic.
• Pressure changes, significant for gases, move the equilibrium towards the side with fewer moles when pressure increases.

Le Chatelier's principle helps chemists predict how reaction conditions affect outcome and balance within a chemical system.