Problem 95 - A Complete Resource Book in Chemistry for JEE Main

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 \(\mathrm{K}_{\mathrm{p}}\) is \(1.44 \times 10^{-5}\) when partial pressure is measured in atmospheres. The corresponding value of \(\mathrm{K}_{c}\), with concentration in mole \(\mathrm{L}^{-1}\), is (a) \(1.44 \times 10^{-5} /(0.082 \times 500)^{-2}\) (b) \(1.44 \times 10^{-5} /(8.314 \times 773)^{-2}\) (c) \(1.44 \times 10^{-5}(0.082 \times 773)^{2}\) (d) \(1.44 \times 10^{-5} /(0.082 \times 773)^{-2}\)

Step-by-Step Solution

Answer

The correct option is (d) \(1.44 \times 10^{-5} /(0.082 \times 773)^{-2}\).

Step 1: Understand the Relationship Between Kp and Kc

The relationship between the equilibrium constant in terms of partial pressures \(K_p\) and the equilibrium constant in terms of concentrations \(K_c\) is given by the equation: \K_p = K_c(RT)^{\Delta n}\, where \- \(R\) is the ideal gas constant (0.0821 L·atm/mol·K or 8.314 J/mol·K), \- \(T\) is the temperature in Kelvin, \- \(\Delta n\) is the change in moles of gas between products and reactants.

Step 2: Calculate the Change in Moles (\(\Delta n\))

For the reaction \(\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)\),\(\Delta n = moles \: of \: products \: - \: moles \: of \: reactants = 2 - (1 + 3) = 2 - 4 = -2\).

Step 3: Convert Temperature to Kelvin

Given that the temperature is \(500^{\circ}C\), convert it to Kelvin: \[T = 500 + 273 = 773 \: K\].

Step 4: Substitute Known Values into the Expression

We have \(\Delta n = -2\), \(R = 0.0821\) L·atm/mol·K for \(K_p\) in atm and \(T = 773\) K. Use the expression: \[K_p = K_c(RT)^{\Delta n}\]or equivalently, \[K_c = \frac{K_p}{(RT)^{\Delta n}}\].

Step 5: Solve for \(K_c\)

Substitute the known values: \[K_c = \frac{1.44 \times 10^{-5}}{(0.0821 \times 773)^{-2}}\].Solve: Calculate \((0.0821 \times 773)^{2}\) and then divide \(1.44 \times 10^{-5}\) by this result.

Step 6: Compare the Options

After solving in step 5, compare it to the given options:(a) \(1.44 \times 10^{-5} /(0.082 \times 500)^{-2}\)(b) \(1.44 \times 10^{-5} /(8.314 \times 773)^{-2}\)(c) \(1.44 \times 10^{-5}(0.082 \times 773)^{2}\)(d) \(1.44 \times 10^{-5} /(0.082 \times 773)^{-2}\).The correct option is (d) because the expression matches \(K_c = \frac{1.44 \times 10^{-5}}{(0.0821 \times 773)^{-2}}\).

Key Concepts

Reversible Reactions

Reversible reactions are a fascinating concept in chemistry where the reactants can transform into products, and those products can reverse back into the original reactants. This type of reaction is represented by a double arrow (\(\rightleftharpoons\)) indicating that both forward and backward reactions are taking place simultaneously. For instance, the synthesis of ammonia from nitrogen (\(\text{N}_2\)) and hydrogen (\(\text{H}_2\)) is a classic example:\[\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)\]In such systems, equilibrium is achieved when the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant over time. Reversible reactions are crucial in industrial chemical processes where conditions can be adjusted to favor the production of desired products.

Kp to Kc Conversion

Understanding how to convert between \(K_p\) and \(K_c\) is essential when dealing with reactions involving gases. \(K_p\) is the equilibrium constant expressed in terms of partial pressures, while \(K_c\) is based on molarity (concentration). The relationship between these constants is given by:\[K_p = K_c(RT)^{\Delta n}\]Here's what each term represents:

• \(K_p\) and \(K_c\) are the equilibrium constants in terms of pressures and concentrations, respectively.
• \(R\) is the ideal gas constant, which can be 0.0821 L·atm/mol·K or 8.314 J/mol·K depending on units.
• \(T\) is the temperature in Kelvin.
• \(\Delta n\) is the change in moles of gas (products minus reactants).

To convert from \(K_p\) to \(K_c\), use the equation:\[K_c = \frac{K_p}{(RT)^{\Delta n}}\]This is a crucial step in chemical calculations involving gases, as it allows chemists to translate pressure-related data into concentration terms, facilitating more straightforward interpretations of reaction conditions.

Ideal Gas Law

The ideal gas law is a fundamental principle that relates the pressure, volume, temperature, and number of moles of a gas. Expressed mathematically, it is:\[PV = nRT\]Where:

• \(P\) is the pressure of the gas.
• \(V\) is the volume occupied by the gas.
• \(n\) is the number of moles of the gas.
• \(R\) is the ideal gas constant.
• \(T\) is the temperature in Kelvin.

This law assumes gases behave ideally, meaning particles move randomly and occupy negligible space with no intermolecular forces. This provides a very close approximation for real gases under low pressure and high temperature conditions. The ideal gas law is helpful in deriving the values needed in \(K_p\) to \(K_c\) conversions, as it helps in calculating changes in moles which is a component in the \(\Delta n\) term.

Temperature Conversion to Kelvin

In scientific calculations, temperatures are often required in Kelvin rather than degrees Celsius. This conversion is crucial because temperature in \(K\) is absolute and ensures consistency in equations like the ideal gas law.To convert Celsius to Kelvin, add 273.15:\[T_{K} = T_{^{\circ}C} + 273.15\]In the given problem, where the temperature is specified as \(500^{\circ}C\), the conversion involves simplifying the equation by adding 273:\[T = 500 + 273 = 773 \: K\]This conversion helps in maintaining uniformity when using the gas constant \(R\) in calculations, which itself is defined for Kelvin scale operations. Always remember this simple addition rule to avoid miscalculations in scenarios involving temperature-dependent equations.