JEE MAIN - Chemistry (2006 - No. 6)
The equilibrium constant for the reaction
SO3 (g) $$\leftrightharpoons$$ SO2 (g) + $$1 \over 2$$ 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 :
SO3 (g) $$\leftrightharpoons$$ SO2 (g) + $$1 \over 2$$ 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 :
416
9.8 $$\times$$ 10-2
4.9 $$\times$$ 10-2
2.40 $$\times$$ 10-3
Explanation
$$S{O_3}\left( g \right)\,\rightleftharpoons\,S{O_2}\left( g \right)\,\, + \,\,{1 \over 2}{O_2}\left( g \right)$$
$${K_c} = {{\left[ {S{O_2}} \right]{{\left[ {{O_2}} \right]}^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} \over {\left[ {S{O_3}} \right]}}$$
$$ = 4.9 \times {10^{ - 2}};$$
On taking the square of the above reaction
$${{{{\left[ {S{O_2}} \right]}^2}\left[ {{O_2}} \right]} \over {{{\left[ {S{O_3}} \right]}^2}}}$$
$$ = 24.01 \times {10^{ - 4}}$$
now $$K{'_C}$$
for $$\,\,2S{O_2}\left( g \right)\,\, + \,\,{O_2}\left( g \right)\,\rightleftharpoons\,2S{O_3}$$
$$ = {{{{\left[ {S{O_3}} \right]}^2}} \over {{{\left[ {S{O_2}} \right]}^2}\left[ {{O_2}} \right]}}$$
$$ = {1 \over {24.01 \times {{10}^{ - 4}}}}$$
$$ = 416$$
$${K_c} = {{\left[ {S{O_2}} \right]{{\left[ {{O_2}} \right]}^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} \over {\left[ {S{O_3}} \right]}}$$
$$ = 4.9 \times {10^{ - 2}};$$
On taking the square of the above reaction
$${{{{\left[ {S{O_2}} \right]}^2}\left[ {{O_2}} \right]} \over {{{\left[ {S{O_3}} \right]}^2}}}$$
$$ = 24.01 \times {10^{ - 4}}$$
now $$K{'_C}$$
for $$\,\,2S{O_2}\left( g \right)\,\, + \,\,{O_2}\left( g \right)\,\rightleftharpoons\,2S{O_3}$$
$$ = {{{{\left[ {S{O_3}} \right]}^2}} \over {{{\left[ {S{O_2}} \right]}^2}\left[ {{O_2}} \right]}}$$
$$ = {1 \over {24.01 \times {{10}^{ - 4}}}}$$
$$ = 416$$
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