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\(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) in the above reaction \(\mathrm{K}_{\mathrm{p}}\) and \(\mathrm{K}_{c}\) are related as (a) \(\mathrm{K}=\mathrm{K}_{\mathrm{c}} \times(\mathrm{RT})\) (b) \(K_{B}=K_{c} \times(R T)^{-1}\) (c) \(\mathrm{K}_{\mathrm{c}}=\mathrm{K}_{\mathrm{p}} \times(\mathrm{RT})^{2}\) (d) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} \times(\mathrm{RT})^{-2}\)

Short Answer

Expert verified
The correct relation is option (b): \(K_{p}=K_{c} \times(R T)^{-1}\).

Step by step solution

01

Understand the Reaction Formula

The given reaction is \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\). Historically, \(K_p\) and \(K_c\) are used to denote equilibrium constants based on pressures and concentrations, respectively.
02

Define Relationship between \(K_p\) and \(K_c\)

The general formula relating \(K_p\) and \(K_c\) is given by:\[K_p = K_c \times (RT)^{\Delta n}\]where \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin, and \(\Delta n\) is the change in moles of gas (products - reactants).
03

Calculate \(\Delta n\) for the Reaction

In our reaction:- Moles of gaseous products = 2 (from \(2 \mathrm{SO}_{3}\))- Moles of gaseous reactants = 2 (\(2 \mathrm{SO}_2) + 1 (\mathrm{O}_2) = 3\)Therefore, \(\Delta n = 2 - 3 = -1\).
04

Apply \(\Delta n\) in Formula

Substitute \(\Delta n = -1\) into the formula:\[K_p = K_c \times (RT)^{-1}\]This shows that \(K_p\) is related to \(K_c\) by multiplying with \((RT)^{-1}\).
05

Match with Given Options

The relationship \(K_p = K_c \times (RT)^{-1}\) matches with option (b), which states\(K_{p}=K_{c} \times(R T)^{-1}\).

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Key Concepts

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

Relation between Kp and Kc
In the world of chemical reactions, especially those involving gases, we often encounter two different expressions for the equilibrium constant: \(K_p\) and \(K_c\). The difference lies in the units used. \(K_p\) is based on the partial pressures of the gaseous components, while \(K_c\) is based on molar concentrations.
To connect these two expressions, scientists use the relation \[K_p = K_c \times (RT)^{\Delta n}\] where \(R\) is the universal gas constant, \(T\) is temperature in Kelvin, and \(\Delta n\) represents the change in moles of gas during the reaction.
This important equation helps us predict how changes in conditions impact the equilibrium constants, making it a pivotal tool in gas reaction analysis.
Equilibrium in Gaseous Reactions
Reactions involving gases not only have balance in their equations but also in their behavior when allowed to proceed to equilibrium. Gaseous equilibrium plays a significant role in understanding reaction dynamics.
In these types of reactions, balanced conditions are achieved when the rate of the forward reaction equals the rate of the reverse reaction, and concentrations of reactants and products remain constant.
For gaseous reactions, pressures and concentrations can change, but at equilibrium, the ratio described by the equilibrium constant \(K\) remains stable. This allows us to make predictions about system behavior under various conditions.
Change in Moles (Delta n)
The term \(\Delta n\) is essential when connecting \(K_p\) and \(K_c\) for gaseous reactions. It represents the change in moles of gas from reactants to products, given by \(\Delta n = n_{\text{products}} - n_{\text{reactants}}\).
In the reaction \(2 \mathrm{SO}_2 (\mathrm{g}) + \mathrm{O}_2 (\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_3 (\mathrm{g})\), \(n_{\text{products}}\) is 2 (as indicated by \(2 \mathrm{SO}_3\)), and \(n_{\text{reactants}}\) is 3 (as indicated by the sum of \(2 \mathrm{SO}_2\) and \(\mathrm{O}_2\)). Thus, \(\Delta n\) is \(-1\).
This simple calculation can have a significant effect on the relation between equilibrium constants, as negative \(\Delta n\) signifies a decrease in the number of moles from reactants to products, influencing how \(K_p\) and \(K_c\) relate.
Universal Gas Constant R
The universal gas constant \(R\) is a cornerstone in chemistry and physics. It connects the energy scale to temperature and relates pressure to volume and temperature for gasses.
  • The value of \(R\) is 8.314 J/mol·K when using SI units.
  • In the equation \(K_p = K_c \times (RT)^{\Delta n}\), \(R\) adjusts the equilibrium constant from concentration units to pressure units.
  • R's consistency across calculations provides a standard, simplifying the translation of theoretical calculations into practical, real-world applications.
Understanding \(R\) allows students and professionals to work effectively with gaseous reactions, predicting behavior under various thermodynamic conditions.

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Most popular questions from this chapter

Equilibrium constant for the reaction \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g})\) is 81. If the velocity constant of the forward reaction is \(162 \mathrm{~L}\) \(\mathrm{mol}^{-1} \sec ^{-1}\), what is the velocity constant (in \(\mathrm{L} \mathrm{mol}^{-1}\) \(\sec ^{-1}\) ) for the backward reaction? (a) 13122 (b) 2 (c) 261 (d) 243

At \(25^{\circ} \mathrm{C}\), the standard emf of a cell having reaction involving two electron exchange is found to be \(0.295 \mathrm{~V} .\) The equilibrium constant of the reaction is approximately (a) \(9.50 \times 10^{9}\) (b) \(1 \times 10^{10}\) (c) 10 (d) \(9.51 \times 10^{\top}\)

If \(\mathrm{K}_{1}\) and \(\mathrm{K}_{2}\) are the respective equilibrium constants for the two reactions, \(\mathrm{XeF}_{6}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{XeOF}_{4}(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g})\) \(\mathrm{XeO}_{4}(\mathrm{~g})+\mathrm{XeF}_{6}(\mathrm{~g}) \rightleftharpoons \mathrm{XeOF}_{4}(\mathrm{~g})+\mathrm{XeO}_{3} \mathrm{~F}_{2}(\mathrm{~g})\) Then equilibrium constant of the reaction \(\mathrm{XeO}_{4}\) (g) + \(2 \mathrm{HF}(\mathrm{g}) \rightleftharpoons \mathrm{XeO}_{3} \mathrm{~F}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) will be (a) \(\mathrm{K}_{1} /\left(\mathrm{K}_{2}\right)^{2}\) (b) \(\mathrm{K}_{1} \cdot \mathrm{K}_{2}\) (c) \(\mathrm{K}_{1} / \mathrm{K}_{2}\) (d) \(\mathrm{K}_{2} / \mathrm{K}_{\mathrm{t}}\)

In a \(0.5\) litre capacity vessel, \(\mathrm{CO}\) and \(\mathrm{Cl}\), are mixed to form \(\mathrm{COCl}_{2} .\) At equilibrium, it contains \(0.2\) mole of \(\mathrm{COCl}_{2}\) and \(0.1\) mole each of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\). The equilibrium constant \(\left(\mathrm{K}_{c}\right)\) for reaction \(\mathrm{CO}+\mathrm{Cl}_{2} \rightleftharpoons \mathrm{COCl}_{2}\) is (a) 15 (b) 5 (c) 20 (d) 10

Match the following Column-I (a) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\) (b) \(\mathrm{CaCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\) (c) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) (d) \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) Column-II (p) Unaffected by inert gas addition at constant volume (q) Forward shift by rise in pressure (r) Unaffected by increase in pressure (s) Backward shift by rise in pressure (t) reaction has \(\Delta n_{g}>0\)

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