For the reaction A_{(g)} + 2B_{(g)} rightleft | vedclass

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(B) Initial moles: $A = 1 \ mol$,$B = 1.5 \ mol$,$C = 0 \ mol$. Volume $V = 2 \ L$. Initial concentrations: $[A]_0 = 1/2 = 0.5 \ M$,$[B]_0 = 1.5/2 = 0

Vedclass · Accepted Answer

$2.36$

Vedclass · Answer

(B) Initial moles: $A = 1 \ mol$,$B = 1.5 \ mol$,$C = 0 \ mol$. Volume $V = 2 \ L$. Initial concentrations: $[A]_0 = 1/2 = 0.5 \ M$,$[B]_0 = 1.5/2 = 0.75 \ M$. At equilibrium,$[C] = 0.35 \ M$. From the stoichiometry $A + 2B ightleftharpoons 2C$,if $2x = 0.35$,then $x = 0.175 \ M$. Equilibrium concentrations: $[A] = 0.5 - x = 0.5 - 0.175 = 0.325 \ M$. $[B] = 0.75 - 2x = 0.75 - 0.35 = 0.4 \ M$. $[C] = 0.35 \ M$. $K_c = \frac{[C]^2}{[A][B]^2} = \frac{(0.35)^2}{(0.325)(0.4)^2} = \frac{0.1225}{0.325 	imes 0.16} = \frac{0.1225}{0.052} \approx 2.3557 \approx 2.36 \ M^{-1}$.

For the reaction $A_{(g)} + 2B_{(g)} \rightleftharpoons 2C_{(g)}$,$1 \ mol$ of $A$ and $1.5 \ mol$ of $B$ are taken in a $2 \ L$ vessel. If the concentration of $C$ at equilibrium is $0.35 \ M$,then the equilibrium constant $K_c$ of the reaction will be ....... $M^{-1}$.

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