2 SO2( g)+O2( g) ⇌2 SO3( g), in the above reaction Kp and Kc are related as (a) Kp=Kc ×(RT) (b)

Name: Problem 9, Chapter 7: A Complete Resource Book in Chemistry for JEE Main 2018
Uploaded: 2021-09-28T11:55:22-08:00
Duration: 1 min 44 s
Channel: Rajendra Kumar
Description: 2 SO2( g)+O2( g) ⇌2 SO3( g), in the above reaction Kp and Kc are related as (a) Kp=Kc ×(RT) (b) Kp=Kc ×(RT)^-1 (c) Kc=Kp ×(RT)^2 (d) Kp=Kc ×(RT)^-2

step 1 Hello to text. So in this question we have to find relation between KP and KC for the following

Question

$2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g})$, in the above reaction $\mathrm{K}_{\mathrm{p}}$ and $\mathrm{K}_{\mathrm{c}}$ are related as (a) $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} \times(\mathrm{RT})$ (b) $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} \times(\mathrm{RT})^{-1}$ (c) $\mathrm{K}_{\mathrm{c}}=\mathrm{K}_{\mathrm{p}} \times(\mathrm{RT})^2$ (d) $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} \times(\mathrm{RT})^{-2}$

    $2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g})$, in the above reaction $\mathrm{K}_{\mathrm{p}}$ and $\mathrm{K}_{\mathrm{c}}$ are related as
(a) $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} \times(\mathrm{RT})$
(b) $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} \times(\mathrm{RT})^{-1}$
(c) $\mathrm{K}_{\mathrm{c}}=\mathrm{K}_{\mathrm{p}} \times(\mathrm{RT})^2$
(d) $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} \times(\mathrm{RT})^{-2}$

A Complete Resource Book in Chemistry for JEE Main 2018

A.K. Singhal 3rd Edition

Chapter 7, Problem 9

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Solved by Expert Rajendra Kumar

09/28/2021

Step 1

The given reaction is: \[ 2 \mathrm{SO}_2(\mathrm{~g}) + \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g}) \] We need to relate the equilibrium constants $ K_p $ and $ K_c $. Show more…

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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 $\mathrm{K}_{\mathrm{c}}$ are related as (a) $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} \times(\mathrm{RT})$ (b) $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} \times(\mathrm{RT})^{-1}$ (c) $\mathrm{K}_{\mathrm{c}}=\mathrm{K}_{\mathrm{p}} \times(\mathrm{RT})^2$ (d) $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} \times(\mathrm{RT})^{-2}$

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Key Concepts

Conversion between Kc and Kp

The relationship between Kc and Kp is given by the equation Kp = Kc(RT)^(?n), where R is the universal gas constant, T is the temperature in Kelvin, and ?n is the net change in the number of moles of gas during the reaction. This relation bridges the gap between concentration and pressure measurements in gas-phase equilibria.

Change in Moles of Gas (?n)

?n represents the difference between the total moles of gaseous products and the total moles of gaseous reactants. Determining ?n is crucial when applying the Kp to Kc conversion formula, as it directly influences the exponent of the (RT) term in the relationship.

Equilibrium Constants

Equilibrium constants quantitatively describe the ratio of products to reactants at equilibrium. In the context of gases, two common forms are Kc, which is based on concentration, and Kp, which is based on partial pressure.

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