Equilibrium Ques 83
83. The reaction $2 A(g)+B(g) \rightarrow 3 C(g)+D(g)$ is begun with the concentrations of $A$ and $B$ both at an initial value of $1.00 M$. When equilibrium is reached, the concentration of $D$ is measured and found to be $0.25 M$. The value for the equilibrium constant for this reaction is given by the expression
[2010]
(a) $[(0.75)^{3}(0.25)] \div[(0.75)^{2}(0.25)]$
(b) $[(0.75)^{3}(0.25)] \div[(1.00)^{2}(1.00)]$
(c) $[(0.75)^{3}(0.25)] \div[(0.50)^{2}(0.75)]$
(d) $[(0.75)^{3}(0.25)] \div[(0.50)^{2}(0.25)]$
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Solution:
- (c)
Mole ratio
$ 2 A _{(g)}+B _{(g)} \rightarrow 3 C _{(g)}+D _{(g)} $
$\begin{matrix} \text{ Molar concentration } 1 & 1 & 0 & 0\end{matrix} $ at $t=0$
$\begin{matrix} \text{ Molar } & 0.50 & 0.75 & 0.75 & 0.25\end{matrix} $
concentration at equilibrium
$K_c=\frac{[C]^{3}[D]}{[A]^{2}[B]}=\frac{(0.75)^{3}(0.25)}{(0.50)^{2}(0.75)}$
BETA
