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For reversible reaction,$X\;\;\overset{\overset{k_1}{\rightleftharpoons}}{k_2}\;\;Y$.$k_1=2hr^{-1}$ and $\large\frac{[X_{eq}]}{[Y_{eq}]}$$=6$.If $[X_0]=2M,[Y_0]=0M$.What will be the the concentration of Y after 1hr?

$\begin{array}{1 1}(a)\;0.285M&(b)\;0.135M\\(c)\;0.295M&(d)\;0.564M\end{array}$

1 Answer

$X\;\;\overset{\overset{k_1}{\rightleftharpoons}}{k_2}\;\;Y$
At $t=0 X\Rightarrow 2M$
At $X=1hr\;X\Rightarrow 2-x,Y\Rightarrow x$
At $t=t_{eq} X\Rightarrow 2-x_{eq},Y\Rightarrow x_{eq}$
Given :
$k_{eq}=\large\frac{[Y_{eq}]}{[X_{eq}]}=\frac{1}{6}=\frac{K_1}{K_2}$
$\Rightarrow K_2=12hr^{-1}$
Now,
$\large\frac{x_{eq}}{2-x_{eq}}=\frac{1}{6}$
$k=\large\frac{1}{t}$$ln\large\frac{[A_0]}{[A_t]}$
$\Rightarrow (k_1+k_2)t=ln\big(\large\frac{x_{eq}}{x_{eq}-x}\big)$
$(14)(1)=ln\big(\large\frac{2/7}{2/7-x}\big)$
$\Rightarrow x=2/7-\large\frac{2/7}{e^{14}}$$=0.285$
Hence (a) is the correct answer.
answered Dec 19, 2013 by sreemathi.v
 

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