The equilibrium constants at 298 K for a reaction $\mathrm{A}+\mathrm{B}\rightleftharpoons \mathrm{C}+\mathrm{D}$ is 100. If the initial concentration of all the four species were 1 M each, then equilibrium concentration of D (in mol ${\mathrm{L}}^{-1}$ ) will be

• A

  1.182
• B

  0.182
• C

  0.818
• Correct answer
  D

  1.818

Solution

$\mathrm{A}+\mathrm{B}\rightleftharpoons \mathrm{C}+\mathrm{D}$ $\mathrm{K}=100$

$\mathrm{Q}=\frac{1\times 1}{1\times 1}=1$

Hence $\mathrm{Q}<\mathrm{K}$ so reaction moves in forward reaction



$\frac{(1+\mathrm{x}{)}^2}{{\left(1-\mathrm{x}\right)}^2}=100$

$\frac{1+\mathrm{x}}{1-\mathrm{x}}=10$

$1+\mathrm{x}=10-10\mathrm{x}$

$\mathrm{x}=\frac{9}{11}$

So, $=1+\mathrm{x}=1+\frac{9}{11}=1.818$ M

Hence, the correct answer is option: 'D'.