In a reaction,A + B rightleftharpoons C + D,4 | vedclass

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(A) The reaction is $A + B \rightleftharpoons C + D$. Initial moles: $A = 1, B = 1, C = 0, D = 0$. At equilibrium,$40 \%$ of $B$ has reacted,so the am

Vedclass · Accepted Answer

$0.44$

Vedclass · Answer

(A) The reaction is $A + B ightleftharpoons C + D$. Initial moles: $A = 1, B = 1, C = 0, D = 0$. At equilibrium,$40 \%$ of $B$ has reacted,so the amount reacted is $1 	imes 0.4 = 0.4 \ mole$. Equilibrium moles: $A = (1 - 0.4) = 0.6, B = (1 - 0.4) = 0.6, C = 0.4, D = 0.4$. Since the volume is $10 \ L$,the concentrations are $[A] = 0.6/10, [B] = 0.6/10, [C] = 0.4/10, [D] = 0.4/10$. $K_C = \frac{[C][D]}{[A][B]} = \frac{(0.4/10) 	imes (0.4/10)}{(0.6/10) 	imes (0.6/10)} = \frac{0.4 	imes 0.4}{0.6 	imes 0.6} = \frac{0.16}{0.36} = 0.44$.

In a reaction,$A + B \rightleftharpoons C + D$,$40 \%$ of $B$ has reacted at equilibrium,when $1 \ mole$ of $A$ was heated with $1 \ mole$ of $B$ in a $10 \ L$ closed vessel. The value of $K_C$ is

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