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

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Question

(C) For the reaction $2A_{(g)} + B_{(g)} \rightleftharpoons 3C_{(g)} + D_{(g)}$: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

(C) For the reaction $2A_{(g)} + B_{(g)} \rightleftharpoons 3C_{(g)} + D_{(g)}$: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,the changes are: $\Delta[A] = -2(0.25) = -0.50 \ M$,$\Delta[B] = -1(0.25) = -0.25 \ M$,$\Delta[C] = +3(0.25) = +0.75 \ M$.Equilibrium concentrations: $[A] = 1.00 - 0.50 = 0.50 \ M$,$[B] = 1.00 - 0.25 = 0.75 \ M$,$[C] = 0.75 \ M$,$[D] = 0.25 \ M$.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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