The equilibrium constant for the reaction N_{ | vedclass

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(B) For the reaction $N_{2(g)} + O_{2(g)} ightleftharpoons 2 NO_{(g)}$,the equilibrium constant is $K_C = 4 	imes 10^{-4}$.For the reaction $NO_{(g

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(B) For the reaction $N_{2(g)} + O_{2(g)} ightleftharpoons 2 NO_{(g)}$,the equilibrium constant is $K_C = 4 	imes 10^{-4}$.For the reaction $NO_{(g)} ightleftharpoons \frac{1}{2} N_{2(g)} + \frac{1}{2} O_{2(g)}$,the reaction is the reverse of the original reaction and multiplied by a factor of $\frac{1}{2}$.Therefore,the new equilibrium constant $K_C'$ is given by $K_C' = \frac{1}{\sqrt{K_C}}$.$K_C' = \frac{1}{\sqrt{4 	imes 10^{-4}}} = \frac{1}{2 	imes 10^{-2}} = \frac{1}{0.02} = 50$.

Components	$\%$ of the total cellular mass
$A$. Water	$1$. $10-15$
$B$. Proteins	$2$. $70-90$
$C$. Carbohydrates	$3$. $5-7$
$D$. Lipids	$4$. $3$
$E$. Nucleic acids	$5$. $2$
$F$. Ions	$6$. $1$

The equilibrium constant for the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)}$ at temperature $T$ is $4 \times 10^{-4}$. The value of $K_C$ for the reaction $NO_{(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{1}{2} O_{2(g)}$ at the same temperature is:

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