[desifabalus]

Consider the following reaction:

4NH3 (g) +7O2 (g) <-->2N2O4 (g) + 6H2O (g)

If the initial concentrations of  N2O4 and H2O are 3.60 M. At equilibrium the concentration of H2O is 0.60 M. Calculate the equilibrium concentration of N2O4

Can someone tell me how to solve this, i am lost, thanks!

[xiankai]

"If the initial concentrations of  N2O4 and H2O are 3.60 M."

do u mean both combined, or for both respectively?

anyway, to solve this equation,  convert all given data to moles first.

applying what we know, concentration is mol/dm³, and all gases have a definite constant volume of 24dm³ per mole.

thus, find how many moles of water there were initially, then based on the equation, 4 moles of ammonia react with 7 moles of oxygen to form 2 moles of hydrazine and 6 moles of water.

subtract the moles of water in equilibrium from the moles of water initially, then u will see how many moles of water were converted.

then u calculate the respective moles of hydrazine that must be converted too, and then find subtract that from the initial moles of hydrazine. that gives u the moles of hydrazine in equilibrium. then, apply the gas volume constant and u have your answer!

[AWK]

Information on constant volume or constant pressure and temperatue are missing.
Taking into account the constant volume you can assume volume of 1 liter and use moles or concentration (they are equivalent in this case)
Hence at the equilibrium you have 0.6 mole of water, 2.6 mole of N2O4, 3.5 moles of O2 and 2 moles of NH3, all per 1 liter.