A) \[4\times {{10}^{-4}}\]
B) \[2\times {{10}^{-4}}\]
C) \[{{10}^{-4}}\]
D) \[8\times {{10}^{-4}}\]
Correct Answer: A
Solution :
\[2A(g)\,\,+\,\,B(g)\,\,\,\,C(g)\,\,+\,\,D(g)\] Initial moles 1 1 7 3 At equi. \[2-2x\] \[1-x\] \[7+x\] \[3+x\] \[\therefore \] Due to very high value of K, we can assume that reactant almost gets converted into products, so \[1-x=y\] \[2-2x=2y\,\,\Rightarrow \,\,x\sim 1\] \[\therefore \] \[{{10}^{12}}=\frac{[C]\,[D]}{{{[A]}^{2}}\,[B]}\] \[{{10}^{12}}=\frac{8\times 4}{{{(2y)}^{2}}(y)}\] \[{{y}^{3}}=8\times !{{0}^{-12}}\] \[\Rightarrow \] \[y=2\times {{10}^{-4}}\] \[\therefore \] Equilibrium concentration of \[A=2y\] \[=2\times 2\times {{10}^{-4}}=4\times {{10}^{-4}}\]
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