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 \(\mathrm{K}_{\mathrm{c}}\), 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) \(1.44 \times 10^{-5} /(0.082 \times 773)^{-2}\)

Short answer

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

Step-by-step solution

Understand the Reaction

The reaction is \( \text{N}_2(g) + 3 \text{H}_2(g) \rightleftharpoons 2 \text{NH}_3(g) \). This reaction involves gases, and the equilibrium constant can be expressed in terms of partial pressures (\( K_p \)) or concentrations (\( K_c \)).

Establish the Relationship Between Kp and Kc

The formula relating \( K_p \) and \( K_c \) for a gaseous reaction is: \[ K_p = K_c (RT)^{\Delta n} \] where \( R \) is the universal gas constant (0.0821 L atm K^{-1} mol^{-1}), \( T \) is the temperature in Kelvin, and \( \Delta n \) is the change in moles of gas between products and reactants.

Calculate Delta n (∆n)

For the reaction \( \text{N}_2(g) + 3 \text{H}_2(g) \rightleftharpoons 2 \text{NH}_3(g) \), the change in moles of gases, \( \Delta n \), is calculated as follows: \[ \Delta n = \text{moles of products} - \text{moles of reactants} = 2 - (1+3) = 2 - 4 = -2 \]

Convert Temperature to Kelvin

The given temperature is \( 500^\circ \text{C} \). Convert this to Kelvin by adding 273: \( 500 + 273 = 773 \text{ K} \).

Solve for Kc

Rearrange the formula \( K_p = K_c (RT)^{\Delta n} \) to solve for \( K_c \):\[ K_c = \frac{K_p}{(RT)^{\Delta n}} \]Substitute \( K_p = 1.44 \times 10^{-5} \), \( R = 0.0821 \text{ L atm K^{-1} mol^{-1}} \), \( T = 773 \text{ K} \), and \( \Delta n = -2 \):\[ K_c = \frac{1.44 \times 10^{-5}}{(0.0821 \times 773)^{-2}} \]

Match Expression with Options

Simplify the expression for \( K_c \) and compare with provided options. The expression obtained is:\[ K_c = 1.44 \times 10^{-5} \div (0.0821 \times 773)^{-2} \]Looking at the options, this matches option (d):\( 1.44 \times 10^{-5} /(0.082 \times 773)^{-2} \).

Key concepts

Reversible Reactions

In chemistry, reversible reactions are processes where the reactants convert into products and the products simultaneously convert back into the reactants. This back-and-forth process continues until an equilibrium state is reached.
At this point, the rate of the forward reaction equals the rate of the reverse reaction, resulting in a constant concentration of both reactants and products. It's important to note that equilibrium doesn't mean the reactants and products are equal, just that their ratios remain constant over time.

• Takes place in closed systems where no substances are added or lost to the surroundings.
• Can be influenced by changes in temperature, pressure, and concentration.

The reversible nature of reactions forms the foundation for understanding equilibrium concepts in chemistry, as described in the given textbook exercise.

Equilibrium Constants

Equilibrium constants, denoted as \(K_p\) and \(K_c\), are numerical values that express the ratio of the concentrations of products to reactants at equilibrium. In gaseous reactions, \(K_p\) deals with partial pressures whereas \(K_c\) uses concentrations in mole \(L^{-1}\).
The relationship between \(K_p\) and \(K_c\) for reactions involving gases is given by the equation \[ K_p = K_c (RT)^{\Delta n} \], where \(R\) is the gas constant, \(T\) is the temperature in Kelvin, and \(\Delta n\) represents the difference in the number of moles of products and reactants.

• Useful for predicting the direction of a reaction.
• Serves as an indicator of reaction extent, with larger values suggesting a higher concentration of products at equilibrium.

The solution to the exercise illustrates how these constants interrelate and can be used to derive conditions such as temperature for a specific equilibrium state.

Gaseous Reactions

Gaseous reactions are reactions where all the reactants and products are in the gas phase. These reactions frequently occur under various conditions of pressure and volume, making understanding them vital for applications in real-world scenarios like industrial syntheses.
In the exercise, the reaction \(\text{N}_2(g) + 3 \text{H}_2(g) \rightleftharpoons 2 \text{NH}_3(g)\) is investigated, showcasing a typical gaseous reaction.

• Partial pressures play a crucial role in determining the equilibrium states of gaseous reactions.
• The concepts of \(\Delta n\) and volume are essential in applying the ideal gas law and equilibrium calculations.

This reaction, known as the Haber process, is a key industrial method for producing ammonia, highlighting the importance of optimizing conditions for maximum yield.

Thermodynamics in Chemistry

Thermodynamics in chemistry deals with the study of energy changes, especially focusing on heat exchange and work during chemical reactions. For reversible reactions, understanding thermodynamics is crucial as it governs the conditions at which forward and reverse reactions proceed at the same rate.
The exercise demonstrates applying thermodynamic principles in calculating equilibrium constants, and understanding concepts like \(\Delta n\), the change in moles results from the reaction gauge energy changes.

• Relates to the distribution of energy between reactants and products.
• Informs on the stability of compounds and conditions favorable for reaction passages.

Grasping these concepts ensures one better predicts and manipulates chemical processes, offering insights into how energy and matter interact.