The equilibrium constant K_{c} at 298 K for | vedclass

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(B) The reaction is $A + B \rightleftharpoons C + D$ with $K_{eq} = 100$. Initial concentrations are $[A] = 1 \ M, [B] = 1 \ M, [C] = 1 \ M, [D] = 1 \

Vedclass · Accepted Answer

$182$

Vedclass · Answer

(B) The reaction is $A + B ightleftharpoons C + D$ with $K_{eq} = 100$. Initial concentrations are $[A] = 1 \ M, [B] = 1 \ M, [C] = 1 \ M, [D] = 1 \ M$. Calculate the reaction quotient $Q_{c} = \frac{[C][D]}{[A][B]} = \frac{1 	imes 1}{1 	imes 1} = 1$. Since $Q_{c} Let $x$ be the change in concentration at equilibrium: $[A] = 1-x, [B] = 1-x, [C] = 1+x, [D] = 1+x$. $K_{eq} = \frac{(1+x)(1+x)}{(1-x)(1-x)} = 100$. Taking the square root: $\frac{1+x}{1-x} = 10$. $1+x = 10 - 10x$ $\Rightarrow 11x = 9$ $\Rightarrow x = \frac{9}{11} \approx 0.818$. Equilibrium concentration of $D = 1 + x = 1 + \frac{9}{11} = \frac{20}{11} \approx 1.8181 \ M$. Expressing in terms of $10^{-2} \ M$: $1.8181 	imes 10^{0} = 181.81 	imes 10^{-2} \ M$. Rounding to the nearest integer,we get $182 	imes 10^{-2} \ M$.

The equilibrium constant $K_{c}$ at $298 \ K$ for the reaction $A + B \rightleftharpoons C + D$ is $100$. Starting with an equimolar solution with concentrations of $A$,$B$,$C$ and $D$ all equal to $1 \ M$,the equilibrium concentration of $D$ is $....... \times 10^{-2} \ M$. (Nearest integer)

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