For the reaction 2NO_{2(g)} rightleftharpoons | vedclass

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(C) Given the first reaction: $2NO_{2(g)} ightleftharpoons 2NO_{(g)} + O_{2(g)}$ with $K_{c1} = 1.8 	imes 10^{-6}$.For the second reaction: $NO_{(g

Vedclass · Accepted Answer

$7.5 	imes 10^{2}$

Vedclass · Answer

(C) Given the first reaction: $2NO_{2(g)} ightleftharpoons 2NO_{(g)} + O_{2(g)}$ with $K_{c1} = 1.8 	imes 10^{-6}$.For the second reaction: $NO_{(g)} + \frac{1}{2}O_{2(g)} ightleftharpoons NO_{2(g)}$,the equilibrium constant is $K_{c2}$.The second reaction is the reverse of the first reaction multiplied by a factor of $\frac{1}{2}$.Therefore,$K_{c2} = (\frac{1}{K_{c1}})^{1/2} = \frac{1}{\sqrt{K_{c1}}}$.$K_{c2} = \frac{1}{\sqrt{1.8 	imes 10^{-6}}} = \frac{1}{1.3416 	imes 10^{-3}} \approx 7.45 	imes 10^2 \approx 7.5 	imes 10^2$.

$X=x$	$P(X=x)$
$0$	$0.4$
$1$	$0.3$
$2$	$0.1$
$3$	$0.1$
$4$	$0.1$

For the reaction $2NO_{2(g)} \rightleftharpoons 2NO_{(g)} + O_{2(g)}$,$K_c = 1.8 \times 10^{-6}$ at $185 \ ^oC$. At $185 \ ^oC$,the $K_c$ for the reaction $NO_{(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons NO_{2(g)}$ is:

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