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What Is Kc For the Following Equilibrium When the Equilibrium Concentration of Each Substance Is: [So2]= 0.60 M, [O2] = 0.82 M and [So3] = 1.90 M - CBSE (Science) Class 11 - Chemistry

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Question

What is Kc for the following equilibrium when the equilibrium concentration of each substance is: [SO2]= 0.60 M, [O2] = 0.82 M and [SO3] = 1.90 M ?

2SO2(g) + O2(g) ⇌2SO3(g)` 

Solution 1

The equilibrium constant (Kc) for the give reaction is:

`K_c = [SO_3]^2/([SO_2]^2[O_2])`

`= ((1.90)^2 M^2)/((0.60)^2 (0.821)M^3)`

`= 12.239 M^(-1) ("approximately")`

Hence, Kcfor the equilibrium is `12.239 M^(-1)`

Solution 2

`2SO_2(g) + O_2(g)` ⇌ `2 SO_3 (g)`

`K_c = [SO_3]^2/[SO_2]^2[O_2] = ((1.9 M) xx (1.9 M))/((0.6 M) xx (0.6 M) xx (0.82 M))`

`= 12.229 M^(-1) = 12.229 Lmol^(-1)`

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Solution What Is Kc For the Following Equilibrium When the Equilibrium Concentration of Each Substance Is: [So2]= 0.60 M, [O2] = 0.82 M and [So3] = 1.90 M Concept: Law of Chemical Equilibrium and Equilibrium Constant.
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