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In the reaction, \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}+\) heat, relationship between \(\mathrm{K}_{\mathrm{P}}\) and \(\mathrm{K}_{\mathrm{c}}\) is: (a) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{-2}\) (b) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{2}\) (c) \(K_{p}=K_{c}(R T)^{-3}\) (d) \(\mathrm{K}_{\mathrm{c}}=\mathrm{K}_{\mathrm{p}}(\mathrm{RT})^{3}\)

Short Answer

Expert verified
Option (a): \(K_{p} = K_{c}(R T)^{-2}\).

Step by step solution

01

Understanding the Relationship between Kp and Kc

For gaseous reactions, the relationship between the equilibrium constants Kp and Kc is given by the equation: \[ K_{p} = K_{c}(R T)^{\Delta n} \]where: - \( R \) is the universal gas constant, - \( T \) is the temperature in Kelvin, - \( \Delta n \) is the change in the number of moles of gas (moles of gaseous products minus moles of gaseous reactants).
02

Determine Δn for the Given Reaction

The reaction is: \[ \mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) + \text{heat} \]Calculate \( \Delta n \) as follows:- Moles of gaseous products = 2 (from \( 2 \mathrm{NH}_{3} \))- Moles of gaseous reactants = 1 (from \( \mathrm{N}_{2} \)) + 3 (from \( 3 \mathrm{H}_{2} \)) = 4Hence, \( \Delta n = 2 - 4 = -2 \).
03

Apply Δn to the Kp and Kc Relationship

Substitute \( \Delta n = -2 \) into the relation.\[ K_{p} = K_{c}(R T)^{-2} \]This shows how the equilibrium constants are related based on the change in moles of gases in the reaction.
04

Identify the Correct Option

From the calculation, the correct expression for the relationship between \( K_{p} \) and \( K_{c} \) matches option (a):\[ K_{p} = K_{c}(R T)^{-2} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kp and Kc Relationship
In chemical reactions involving gases, equilibrium constants are used to express the concentrations of reactants and products at equilibrium. There are two types of equilibrium constants when dealing with gaseous reactions: \( K_c \) and \( K_p \). - **\( K_c \)**: It represents the equilibrium constant in terms of molar concentrations of reactants and products.- **\( K_p \)**: It represents the equilibrium constant in terms of partial pressures of the gases involved.The relationship between \( K_p \) and \( K_c \) is given by the equation:\[K_{p} = K_{c}(R T)^{\Delta n} \]where:- \( R \) is the universal gas constant,- \( T \) is the temperature in Kelvin,- \( \Delta n \) is the change in moles of gas (number of moles of gaseous products minus number of moles of gaseous reactants). When \( \Delta n \) is positive, the value of \( K_p \) will be larger than \( K_c \), and when \( \Delta n \) is negative, \( K_p \) will be smaller. This formula helps us to convert one equilibrium constant into another, depending on what is known and needed for solving specific equilibrium problems.
Chemical Equilibrium
Chemical equilibrium is a state in a chemical reaction where the concentrations of reactants and products remain constant over time. This occurs when the rate of the forward reaction is equal to the rate of the reverse reaction.At equilibrium:- No net change in concentration of reactants and products.- Reactions do not stop; they continue to occur but at the same rate in both directions.Equilibrium can occur in closed systems where no new reactants are added or products removed. The equilibrium state is characterized by equilibrium constants \( K_c \) or \( K_p \), which quantitatively measure the ratio of products to reactants. Moreover, changing external conditions such as pressure or temperature can shift the position of equilibrium as per Le Chatelier's Principle. This principle states that any change in the system's conditions will cause the equilibrium to shift in the direction that counteracts the change.
Gas Reactions
Gas reactions involve reactants and products in their gaseous state. The behavior of gases in chemical reactions often follows the ideal gas laws, making it easier to predict and calculate equilibrium conditions. In a gas-phase reaction, the pressure exerted by each component is crucial as it dictates how the system approaches equilibrium. The effect of volume on gas reactions can lead to shifts in reaction direction as per Le Chatelier's Principle; if volume decreases, pressure increases favoring the side with fewer moles of gas. Understanding gas reactions involves recognizing factors such as temperature and pressure and how they influence the balance of chemical reactions. This helps in predicting the outcome or the extent to which reactants convert to products under given conditions.
Mole Concept
The mole concept is a fundamental principle in chemistry that provides a bridge between the atomic scale and laboratory measurements. It defines the amount of substance as containing the same number of entities (atoms, ions, molecules) as there are in 12 grams of carbon-12.One mole is equivalent to Avogadro's number, approximately \( 6.022 \times 10^{23} \) entities. This powerful concept allows scientists to count atoms by weighing macroscopic quantities and facilitates calculations involving stoichiometry.In reactions, the mole concept allows you to calculate how much product forms from a given amount of reactants or how much reactant is needed to form a desired amount of product. It plays a crucial role in balancing chemical equations and understanding proportions in chemical reactions.

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Most popular questions from this chapter

An amount of solid \(\mathrm{NH}_{4}\) HS in placed in a flask already containing ammonia gas at a certain temperature and \(0.50\) atm pressure. Ammonium hydrogen sulphide decomposes to yield \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{~S}\) gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure is the flask rises to \(0.84 \mathrm{~atm}\), the equilibrium constant for \(\mathrm{NH}_{4} \mathrm{HS}\) decomposition at this temperature is: (a) \(0.30\) (b) \(0.18\) (c) \(0.17\) (d) \(0.11\)

The exothermic formation of \(\mathrm{ClF}_{3}\) is represented by the equation: \(\mathrm{Cl}_{2}(\mathrm{~g})+3 \mathrm{~F}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{ClF}_{3}(\mathrm{~g}) ; \Delta \mathrm{H}=-329 \mathrm{~kJ}\). Which of the following will increase the quantity of \(\mathrm{CIF}_{3}\) in an equilibrium mixture of \(\mathrm{Cl}_{2}, \mathrm{~F}_{2}\) and \(\mathrm{ClF}_{3}\) ? (a) Increasing the temperature (b) Removing \(\mathrm{Cl}_{2}\)

For the hypothetical reactions, the equilibrium constant \((\mathrm{K})\) values are given: \(\mathrm{A} \rightleftharpoons \mathrm{B} \mathrm{K}_{1}=2\) \(\mathrm{B} \rightleftharpoons \mathrm{C} \mathrm{K}_{2}^{1}=4\) \(\mathrm{C} \rightleftharpoons \mathrm{D} \mathrm{K}_{3}^{2}=3\) The equilibrium constant \((\mathrm{K})\) for the reaction: \(\mathrm{A} \rightleftharpoons \mathrm{D}\) is (a) 3 (b) 6 (c) 12 (d) 24

For the following three reactions \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\), equilibrium constants are given: (a) \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) ; \mathrm{K}_{1}\) (b) \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g}) ; \mathrm{K}_{2}\) (c) \(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2}(\mathrm{~g}) ; \mathrm{K}_{3}\) Which of the following relation is correct? (a) \(\mathrm{K}_{1} \sqrt{\mathrm{K}_{2}}=\mathrm{K}_{2}\) (b) \(\mathrm{K}_{2} \mathrm{~K}_{3}=\mathrm{K}_{1}\) (c) \(\mathrm{K}_{3}=\mathrm{K}_{1} \mathrm{~K}_{2}\) (d) \(\mathrm{K}_{3} \cdot \mathrm{K}_{2}^{3}=\mathrm{K}_{1}^{2}\)

In a reaction \(\mathrm{A}_{2}(\mathrm{~g})+4 \mathrm{~B}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}_{4}(\mathrm{~g}) ; \Delta \mathrm{H}<0\) The formation of \(\mathrm{AB}_{4}(\mathrm{~g})\) will be favoured by: (a) Low temperature and high pressure (b) High temperature and high pressure (c) Low temperature and low pressure (d) High temperature and low pressure

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