The equilibrium constant at 298 K for the re | vedclass

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(B) The reaction is $A + B \rightleftharpoons C + D$. Initial concentrations: $[A] = 1 \ M, [B] = 1 \ M, [C] = 1 \ M, [D] = 1 \ M$. At equilibrium,let

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$1.818$

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(B) The reaction is $A + B \rightleftharpoons C + D$. Initial concentrations: $[A] = 1 \ M, [B] = 1 \ M, [C] = 1 \ M, [D] = 1 \ M$. At equilibrium,let the change in concentration be $x$. Equilibrium concentrations: $[A] = 1-x, [B] = 1-x, [C] = 1+x, [D] = 1+x$. The equilibrium constant expression is $K_c = \frac{[C][D]}{[A][B]}$. Given $K_c = 100$,we have: $100 = \frac{(1+x)(1+x)}{(1-x)(1-x)}$ Taking the square root on both sides: $10 = \frac{1+x}{1-x}$ $10(1-x) = 1+x$ $10 - 10x = 1 + x$ $9 = 11x$ $x = \frac{9}{11} \approx 0.818 \ M$. The equilibrium concentration of $D$ is $[D] = 1 + x = 1 + 0.818 = 1.818 \ M$.

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

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