1.1 mol of A are mixed with 2.2 mol of B an | vedclass

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(A) The reaction is $A + 2B \rightleftharpoons 2C + D$. Initial moles: $A = 1.1, B = 2.2, C = 0, D = 0$. At equilibrium,let the concentration of $C$ b

Vedclass · Accepted Answer

$0.001$

Vedclass · Answer

(A) The reaction is $A + 2B ightleftharpoons 2C + D$. Initial moles: $A = 1.1, B = 2.2, C = 0, D = 0$. At equilibrium,let the concentration of $C$ be $0.2 \ M$. Since the volume is $1 \ L$,moles of $C = 0.2$. According to the stoichiometry,if $2 \ mol$ of $C$ are formed,$1 \ mol$ of $A$ is consumed and $2 \ mol$ of $B$ are consumed,and $1 \ mol$ of $D$ is formed. For $0.2 \ mol$ of $C$ formed,$0.1 \ mol$ of $A$ is consumed,$0.2 \ mol$ of $B$ is consumed,and $0.1 \ mol$ of $D$ is formed. Equilibrium moles: $A = 1.1 - 0.1 = 1.0 \ mol$,$B = 2.2 - 0.2 = 2.0 \ mol$,$C = 0.2 \ mol$,$D = 0.1 \ mol$. Equilibrium concentrations: $[A] = 1.0 \ M, [B] = 2.0 \ M, [C] = 0.2 \ M, [D] = 0.1 \ M$. $K_c = \frac{[C]^2 [D]}{[A] [B]^2} = \frac{(0.2)^2 (0.1)}{(1.0) (2.0)^2} = \frac{0.04 	imes 0.1}{1.0 	imes 4} = \frac{0.004}{4} = 0.001$.

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