[heen20]

Hey can anyone help me with this question...thanx

a mixture consisting of 0.500mol dm-3 N2 and 0.800mol dm-3 H2 in a reaction vessel reacts and reaches equilibrium.  At equilibrium the concentration of ammonia is 0.150 mol dm-3.  Calculate the value of the equlibrium constant for

N2 (g) + 3H2 (g) ------> 2NH3 (g)

this is what i calculated but not sure if it is right:

Kc =  [0.150]2 / [0.500] [0.800]3  =  8.79 x 10-2

is this right??

thanx for your *delete me*!!

[Yggdrasil]

Your answer is not correct.  0.5mol/dm³ and 0.8mol/dm³ are not the equilibrium concentrations of H₂ and N₂; they are the initial concentrations.  To calculate the final concentrations, you have to take into accound the H₂ and N₂ which reacted to form the NH₃.

Since you end up with 0.150mol/dm³ of ammonia, 0.150 * 1/2 = 0.075mol/dm³ of nitrogen and 0.150 * 3/2 = 0.225mol/dm³ of hydrogen must have reacted.  Therefore, at equilibrium, you have:

conc. N₂ = 0.425 mol/dm³
conc. H₂ = 0.575 mol/dm³
conc. NH₃ = 0.150 mol/dm³

Which gives K_c = 2.78x10^-1 dm⁶/mol²

[jdurg]

Yggdrasil, there must be something wrong with your calculations as well since equillibrium constants are unitless.

[Yggdrasil]

Yggdrasil, there must be something wrong with your calculations as well since equillibrium constants are unitless.

According to my text (Oxtoby, Gills, Nachtrieb.  Principles of Modern Chemistry, 5th ed), empirical equilibrium constants (e.g. K_c) are only dimensionless for reactions for which there is an equal number of molecules on both sides of the reaction.  However, had he said K (the thermodynamic equilibrium constant), you would have been correct.

[heen20]

THANX SO MUCH!!!!!!!!!!!

[heen20]

hi thanx for your help but how did you get these :

conc. N2 = 0.425 mol/dm3
conc. H2 = 0.575 mol/dm3

[Yggdrasil]

During the course of the reaction, you use up 0.075mol/dm³ of nitrogen and 0.225mol/dm³ of hydrogen.  Since you start with 0.5mol/dm³ of nitrogen and use up 0.075mol/dm³, you have 0.500 - 0.075 = 0.425mol/dm³ at equilibrium.  Similarly, you start with 0.8mol/dm³ of hydrogen and use up 0.225mol/dm³, leaving 0.800 - 0.225 = 0.575mol/dm³ of hydrogen at equilibrium.