Chemistry - Chemical Equilibrium Question with Solution | TestHub

$\begin{array}{cccccccc}& 2\text{A}\left(\text{g}\right)& +& \text{B}\left(\text{g}\right)& \rightleftharpoons & \text{C}\left(\text{g}\right)& +& \text{D}\left(\text{g}\right)\\ \text{Initial moles}& 2& & 1& & 7& & 3\\ \text{At equilibrium}& 2-2\text{x}& & 1-\text{x}& & 7+\text{x}& & 3+\text{x}\end{array}$

$\therefore$Due to very high value of$\text{K,}$we can assume that reactant almost gets converted into products, so

$1-\text{x}=\text{y}$

$2-2\text{x}=2\text{y}\Rightarrow \text{x}\sim 1$

$\therefore 10^{12}=\frac{\left[\text{C}\right]\left[\text{D}\right]}{{\left[\text{A}\right]}^2\left[\text{B}\right]}$

$10^{12}=\frac{8\times 4}{{\left(2\text{y}\right)}^2\left(\text{y}\right)}$

${\text{y}}^3=8\times 10^{-12}$

$\Rightarrow \text{y}=2\times 10^{-4}$

$\therefore$Equilibrium concentration of$\text{A}=2\text{y}$

$=2\times 2\times 10^{-4}=4\times 10^{-4}$