Problem 10
Question
In the reaction, \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}+\) heat, relation- ship between \(K_{p}\) and \(K_{c}\) is (a) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{-2}\) (b) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{c}(\mathrm{RT})^{2}\) (c) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{-3}\) (d) \(\mathrm{K}_{\mathrm{c}}=\mathrm{K}_{\mathrm{p}}(\mathrm{RT})^{3}\)
Step-by-Step Solution
Verified Answer
The relationship is (a) \(K_{p} = K_{c}(RT)^{-2}\).
1Step 1: Identify the Reaction Type
The given reaction is \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}+\) heat, which is a gaseous equilibrium reaction where \(\Delta n\) (change in moles of gas) is calculated by subtracting the moles of gaseous reactants from the moles of gaseous products.
2Step 2: Calculate \(\Delta n\)
For the reaction \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\), we calculate \(\Delta n = \text{moles of products} - \text{moles of reactants} = 2 - (1 + 3) = 2 - 4 = -2\).
3Step 3: Use the Relationship Formula
The general relationship between \(K_{p}\) and \(K_{c}\) is given by the formula \(K_{p} = K_{c}(RT)^{\Delta n}\), where \(R\) is the gas constant and \(T\) is the temperature in Kelvin.
4Step 4: Substitute \(\Delta n\) Into the Formula
Substituting \(\Delta n = -2\) into the formula gives us \(K_{p} = K_{c}(RT)^{-2}\).
5Step 5: Match with the Given Options
Compare the derived formula \(K_{p} = K_{c}(RT)^{-2}\) with the provided options to find the matching one: (a) \(\mathrm{K}_{p}= ext{K}_{c}(RT)^{-2}\).
Key Concepts
Gaseous ReactionsChange in Moles (Δn)Kp and Kc Relationship
Gaseous Reactions
Gaseous reactions involve substances that are in the gas phase during the course of the reaction. In these types of reactions, reactants and products are gaseous, allowing them to freely collide and interact with each other. This makes comprehension of gaseous reactions essential in physical chemistry.
- One critical factor in gaseous reactions is the pressure of the system, which can influence the reaction rate and equilibrium.
- These reactions often follow the ideal gas law, which links the pressure, volume, temperature, and amount of gas in the system.
- Equilibria in gaseous reactions are represented using expressions like the equilibrium constant, which depicts the balance between reactants and products.
Change in Moles (Δn)
The concept of change in moles, expressed as \(\Delta n\), is crucial when analyzing gaseous equilibrium reactions. It refers to the difference in the amount of gaseous products and reactants.
When calculating \(\Delta n\), you subtract the total moles of gaseous reactants from the total moles of gaseous products:
When calculating \(\Delta n\), you subtract the total moles of gaseous reactants from the total moles of gaseous products:
- In the reaction \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\), \(\Delta n = 2 - (1 + 3) = -2\).
- \(\Delta n\) can be positive, negative, or zero, and it significantly affects the relationship between different equilibrium constants.
Kp and Kc Relationship
The relationship between \(K_p\) and \(K_c\) links the equilibrium constants for reactions involving gases. \(K_p\) represents the equilibrium constant in terms of partial pressures, while \(K_c\) is in terms of concentrations.
The formula that connects \(K_p\) and \(K_c\) is:
The formula that connects \(K_p\) and \(K_c\) is:
- \(K_p = K_c (RT)^{\Delta n}\), where \(R\) is the ideal gas constant, \(T\) is the temperature in Kelvin, and \(\Delta n\) is the change in moles.
- A negative \(\Delta n\) indicates the reaction will shift toward producing fewer moles of gas, while a positive \(\Delta n\) signifies more moles will be produced.
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