Verified Solution • 2025–26 Session moderate

QJEEM06Ch5673: JEE Main 2006 Chemistry - Class 11 JEEM06Ch5673 Solution: The equilibrium constant for the reaction SO 3 (g) SO 2 (g) + O …

The equilibrium constant for the reaction
SO3 (g) ⇋\leftrightharpoons SO2 (g) + 121 \over 2 O2 (g)
is Kc = 4.9 ×\times 10–2. The value of Kc for the reaction
2SO2 (g) + O2 (g) ⇋\leftrightharpoons 2SO3 (g) will be :

Multiple Choice Options

  • A 416
  • B 9.8 ×\times 10-2
  • C 4.9 ×\times 10-2
  • D 2.40 ×\times 10-3

Step-by-Step Verified Solution

SO3(g) ⇌ SO2(g)  +  12O2(g)S{O_3}\left( g \right)\,\rightleftharpoons\,S{O_2}\left( g \right)\,\, + \,\,{1 \over 2}{O_2}\left( g \right)

Kc=[SO2][O2]1/2[SO3]

=4.9×10−2;= 4.9 \times {10^{ - 2}};

On taking the square of the above reaction

[SO2]2[O2][SO3]2{{{{\left[ {S{O_2}} \right]}^2}\left[ {{O_2}} \right]} \over {{{\left[ {S{O_3}} \right]}^2}}}

=24.01×10−4= 24.01 \times {10^{ - 4}}

now KC′K{'_C}

for   2SO2(g)  +  O2(g) ⇌ 2SO3\,\,2S{O_2}\left( g \right)\,\, + \,\,{O_2}\left( g \right)\,\rightleftharpoons\,2S{O_3}

=[SO3]2[SO2]2[O2]= {{{{\left[ {S{O_3}} \right]}^2}} \over {{{\left[ {S{O_2}} \right]}^2}\left[ {{O_2}} \right]}}

=124.01×10−4= {1 \over {24.01 \times {{10}^{ - 4}}}}

=416= 416

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