Mathman23
Oct20-04, 06:23 PM
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
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