The equations and the values of K are:
2S(s) + 2O₂(g) ⇄ 2SO₂(g) K1 = 2x10¹⁰
2SO₂(g) + O₂(g) ⇄ 2SO₃(g) K2 = 7x10²⁴
Answer:
K3 = 1.4x10³⁵
Explanation:
The equilibrium constant of a global reaction is the multiplication of the constants of the steps reactions. If the equation needed to be multiplied by some constant, the value of the equilibrium constant must be elevated by the same value. If the reaction must be inverted, the value of the equilibrium constant must be inverted (1/K):
2S(s) + 2O₂(g) ⇄ 2SO₂(g) K1 = 2x10¹⁰
2SO₂(g) + O₂(g) ⇄ 2SO₃(g) K2 = 7x10²⁴
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2S(s) + 3O₂(g) + 2SO₂(g) ⇄ 2SO₂(g) + 2SO₃(g)
The substances in both sides are simplified:
2S(s) + 3O₂(g) ⇄ 2SO₃(g)
K3 = K1xK2
K3 = 2x10¹⁰ x 7x10²⁴
K3 = 1.4x10³⁵
The equilibrium constant for the reaction has been 1.4 [tex]\rm \times\;10^3^5[/tex].
The given reactions have been:
[tex]\rm 2\;S\;(s)\;+\;2\;O_2\;(g)\;\leftrightharpoons\;2\;SO_2\;(g)[/tex] ..........Reaction 1
[tex]\rm 2\;SO_2\;(g)\;+\;O_2\;(g)\;\leftrightharpoons\;2\;SO_3\;(g)[/tex].......... Reaction 2
The sum of these two reactions will be:
[tex]\rm 2\;S\;(s)\;+\;3\;O_2\;(g)\;+\;2\;SO_2\;(g)\;\leftrightharpoons\;2\;SO_3\;(g)\;+\;2\;SO_2\;(g)[/tex]
Since [tex]\rm SO_2[/tex] has been in equal concentrations on both sides of reactions, the compound has been eliminated from the final reaction.
[tex]\rm 2\;S\;(s)\;+\;3\;O_2\;(g)\;+\;\leftrightharpoons\;2\;SO_3\;(g)[/tex]
The reaction has been similar to reaction 3. Since reaction 3 has been the sum of reactions 1 and 2, the equilibrium constant of reaction 3 has been the product of the equilibrium constant of the two reactions.
Thus,
K3 = K1 [tex]\times[/tex] K2
K3 = 2 [tex]\rm \times\;10^1^0[/tex] [tex]\times[/tex] 7 [tex]\rm \times\;10^2^4[/tex]
K3 = 1.4 [tex]\rm \times\;10^3^5[/tex]
The equilibrium constant for the reaction has been 1.4 [tex]\rm \times\;10^3^5[/tex].
For more information about the equilibrium constant, refer to the link:
https://brainly.com/question/11684688
