# A mass action expression

1. Oct 20, 2004

### Mathman23

Hi all,

I have the following equilibrium reaction $\mathrm{2NO_{2} \leftrightharpoons N_{2} O_{4}}$ which has a tempeture of
$\mathrm{t = 100 \ ^\circ C.}$
The reactions takes place in a canister with a volume of 0,50 Liters, where the pressure at equilibrium is 1,6 bar.

Futhermore the equilibrium constant is found to be $\mathrm{K_{c} = 5,0 M^{-1}}$

a/ First I write the mass action expression $K_{c} = \frac{\mathrm{[N_{2}O_{4}]}}{\mathrm{[NO_{2}]^2}}$

b/ Second I calculate the amount of mol substance then the reaction is in equilibrium:

$\mathrm{n_{equ}} = \frac{\mathrm{1,6bar \cdot 0,5 L}}{\mathrm{0,0831 \frac{bar \cdot L}{mol \cdot K}\cdot \ {373,0 K}}}= 2,6 \times 10^{-2} \ \mathrm{mol}$

c/ How do I calculate $[\mathrm{NO_{2}}] \ \mathrm{and} \ [\mathrm{N_{2} O_{10}}]$ ??

Many Thanks in advance.

Sincerely
Fred
Denmark

Last edited: Oct 21, 2004
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