JEE MAIN - Chemistry (2019 - 8th April Evening Slot - No. 22)
For the following reactions, equilibrium
constants are given :
S(s) + O2(g) ⇋ SO2(g); K1 = 1052
2S(s) + 3O2(g) ⇋ 2SO3(g); K2 = 10129
The equilibrium constant for the reaction,
2SO2(g) + O2(g) ⇋ 2SO3(g) is :
S(s) + O2(g) ⇋ SO2(g); K1 = 1052
2S(s) + 3O2(g) ⇋ 2SO3(g); K2 = 10129
The equilibrium constant for the reaction,
2SO2(g) + O2(g) ⇋ 2SO3(g) is :
10181
1025
1077
10154
Explanation
S(s) + O2(g) ⇋ SO2(g); K1 = 1052
By reversing the equation, we get
SO2(g); ⇋ S(s) + O2(g) ; $${1 \over {{K_1}}} = {1 \over {{{10}^{52}}}}$$ ..............(1)
2S(s) + 3O2(g) ⇋ 2SO3(g); K2 = 10129 ....................(2)
Performing (2) - 2 $$ \times $$ (1), we get
2SO2(g) + O2(g) ⇋ 2SO3(g)
$$ \therefore $$ Equilibrium constant of this reaction is
= $${{{K_2}} \over {K_1^2}}$$ = $${{{{10}^{129}}} \over {{{\left( {{{10}^{52}}} \right)}^2}}}$$ = 1025
By reversing the equation, we get
SO2(g); ⇋ S(s) + O2(g) ; $${1 \over {{K_1}}} = {1 \over {{{10}^{52}}}}$$ ..............(1)
2S(s) + 3O2(g) ⇋ 2SO3(g); K2 = 10129 ....................(2)
Performing (2) - 2 $$ \times $$ (1), we get
2SO2(g) + O2(g) ⇋ 2SO3(g)
$$ \therefore $$ Equilibrium constant of this reaction is
= $${{{K_2}} \over {K_1^2}}$$ = $${{{{10}^{129}}} \over {{{\left( {{{10}^{52}}} \right)}^2}}}$$ = 1025
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