Problem 10
Question
In the reaction, \(\mathrm{N}_{2}+3 \mathrm{H}_{2}=2 \mathrm{NH}_{3}+\) heat, relationship between \(K_{p}\) and \(K\) is (a) \(\mathrm{K}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{-2}\) (b) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{2}\) (c) \(K_{p}=K_{c}(R T)^{-3}\) (d) \(\mathrm{K}_{\mathrm{c}}=\mathrm{K}_{\mathrm{p}}(\mathrm{RT})^{3}\)
Step-by-Step Solution
Verified Answer
The correct answer is (c) \(K_p = K_c (R T)^{-3}\).
1Step 1: Understand the Reaction Equation
The given reaction is \( \mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g) \). It's crucial to note that this is a gaseous reaction, which allows us to use the concepts of \( K_c \) and \( K_p \).
2Step 2: Calculate Change in Moles (\(\Delta n\))
To find the relationship between \( K_c \) and \( K_p \), use the change in moles. In the reaction, reactants have a total of 4 moles \((1 + 3)\) and products have 2 moles. Thus, \( \Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 2 - 4 = -2 \).
3Step 3: Use the Relation Between \(K_p\) and \(K_c\)
The relationship between \( K_p \) and \( K_c \) is given by the formula \( K_p = K_c (RT)^{\Delta n} \). Substitute \( \Delta n = -2 \) into this equation to get \[ K_p = K_c (RT)^{-2} \].
4Step 4: Match the Equation with Options
Compare the derived equation \( K_p = K_c (RT)^{-2} \) with the given multiple-choice options. The correct match for \( K_p = K_c (RT)^{-2} \) is option (c).
Key Concepts
Equilibrium ConstantGas ReactionsKp and Kc Relationship
Equilibrium Constant
The Equilibrium Constant, often represented as either \( K_c \) or \( K_p \), is a fundamental concept in chemistry that describes the state of balance achieved in a reversible chemical reaction. It is important to understand that this equilibrium is dynamic, meaning that the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant over time.
In any given chemical reaction, the equilibrium constant is calculated based on the concentrations or partial pressures of the reactants and products at equilibrium. For reactions involving gases, the equilibrium constant can be expressed in terms of concentrations (\( K_c \)) or partial pressures (\( K_p \)). Each of these expressions plays a critical role in predicting how a reaction behaves under different conditions.
For example, in a gaseous reaction like \( N_{2}(g) + 3 H_{2}(g) \rightarrow 2 NH_{3}(g) \), the equilibrium constant \( K_c \) would be calculated using the formula: \[ K_c = \frac{[NH_{3}]^2}{[N_2][H_2]^3} \] and \( K_p \) would consider the partial pressures. This allows chemists to determine how far the reaction has proceeded and to predict the concentrations of species at equilibrium.
In any given chemical reaction, the equilibrium constant is calculated based on the concentrations or partial pressures of the reactants and products at equilibrium. For reactions involving gases, the equilibrium constant can be expressed in terms of concentrations (\( K_c \)) or partial pressures (\( K_p \)). Each of these expressions plays a critical role in predicting how a reaction behaves under different conditions.
For example, in a gaseous reaction like \( N_{2}(g) + 3 H_{2}(g) \rightarrow 2 NH_{3}(g) \), the equilibrium constant \( K_c \) would be calculated using the formula: \[ K_c = \frac{[NH_{3}]^2}{[N_2][H_2]^3} \] and \( K_p \) would consider the partial pressures. This allows chemists to determine how far the reaction has proceeded and to predict the concentrations of species at equilibrium.
Gas Reactions
Gas reactions involve reactants and products in the gaseous state. They are significant in studying chemical equilibrium because pressure and temperature can greatly influence the reaction. For a balanced gaseous reaction, like the synthesis of ammonia: \( N_{2}(g) + 3 H_{2}(g) \rightarrow 2 NH_{3}(g) \), understanding the contribution of each component to the overall reaction is crucial.
When dealing with gas reactions,:
When dealing with gas reactions,:
- The behavior and movement of the molecules are dictated by the pressure and temperature conditions within the system.
- These reactions are usually expressed in terms of partial pressures (\( K_p \)) instead of concentrations (\( K_c \)), since gases expand to fill their containers, unlike solids and liquids.
Kp and Kc Relationship
The relationship between \( K_p \) and \( K_c \) is a crucial part of understanding chemical equilibria in systems involving gases. This relationship is defined mathematically to account for the difference in how equilibrium is expressed.
For the ammonia synthesis reaction: \(N_{2}(g) + 3 H_{2}(g) \rightarrow 2 NH_{3}(g) \), \( \Delta n \) is calculated as \(2 - 4 = -2\). Inserting this \( \Delta n \) value into the equation gives:\[ K_p = K_c (RT)^{-2} \] This shows how the two constants are interconnected and helps predict how a system behaves under different pressures and temperatures.
- \( K_c \) is used for concentrations in molarity.
- \( K_p \) refers to partial pressures.
For the ammonia synthesis reaction: \(N_{2}(g) + 3 H_{2}(g) \rightarrow 2 NH_{3}(g) \), \( \Delta n \) is calculated as \(2 - 4 = -2\). Inserting this \( \Delta n \) value into the equation gives:\[ K_p = K_c (RT)^{-2} \] This shows how the two constants are interconnected and helps predict how a system behaves under different pressures and temperatures.
Other exercises in this chapter
Problem 8
For a reversible reaction, the concentration of the reactants are doubled, then the equilibrium constant (a) becomes one-fourth (b) is doubled (c) is halved (d)
View solution Problem 9
\(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) in the above reaction \(\mathrm{K}_{\mathrm{p}}
View solution Problem 11
Which of the following change will shift the reaction in forward direction? \(\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 21(\mathrm{~g})\) Take \(\Delta \ma
View solution Problem 12
A reversible reaction is said to have attained equilibrium, when (a) backward reaction stops (b) both backward and forward reactions take place at equal speed (
View solution