The reaction,2A_{(g)} + B_{(g)} rightleftharp | vedclass

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(B) The reaction is $2A_{(g)} + B_{(g)} \rightleftharpoons 3C_{(g)} + D_{(g)}$.Initial concentrations: $[A] = 1.00 \ M$,$[B] = 1.00 \ M$,$[C] = 0 \ M$

Vedclass · Accepted Answer

$[(0.75)^3(0.25)] \div [(0.50)^2(0.75)]$

Vedclass · Answer

(B) The reaction is $2A_{(g)} + B_{(g)} \rightleftharpoons 3C_{(g)} + D_{(g)}$.Initial concentrations: $[A] = 1.00 \ M$,$[B] = 1.00 \ M$,$[C] = 0 \ M$,$[D] = 0 \ M$.At equilibrium,$[D] = 0.25 \ M$. Since the stoichiometry of $D$ is $1$,the change in concentration for $D$ is $+0.25 \ M$.Using stoichiometry:$[A]_{eq} = 1.00 - 2(0.25) = 0.50 \ M$$[B]_{eq} = 1.00 - 0.25 = 0.75 \ M$$[C]_{eq} = 3(0.25) = 0.75 \ M$$[D]_{eq} = 0.25 \ M$The equilibrium constant expression is $K_c = \frac{[C]^3[D]}{[A]^2[B]}$.Substituting the values: $K_c = \frac{(0.75)^3(0.25)}{(0.50)^2(0.75)}$.

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