Need someone to check my work and help me through the process! PLEASE - equilibriums
Consider the reaction between phosphorous and hydrogen to produce phosphine.
P4 + 6H2 -> 4PH3
0.2 mol of phosphorous and 0.8 mol of hydrogen are placed in an empty 1L vessel. The vessel is sealed and allowed to equilibrate. At equilibrium, 0.2 mol of PH3 are present. Write the equilibrium expression for the reaction and determine the equilibrium constant.
For the equilibrium constant I wrote Kc = Reactants/products = (PH3^4)/(P4)(H2^6)
After finding the equilibrium concentrations for the reactants, my answer is 0.68. I need to know if this is right?
Would the answer be different if it wasn't a 1L vessel? If it was a 2L vessel?
Re: Need someone to check my work and help me through the process! PLEASE - equilibri
Quote:
Originally Posted by
Huggakhan 
Consider the reaction between phosphorous and hydrogen to produce phosphine.
P4 + 6H2 -> 4PH3
0.2 mol of phosphorous and 0.8 mol of hydrogen are placed in an empty 1L vessel. The vessel is sealed and allowed to equilibrate. At equilibrium, 0.2 mol of PH3 are present. Write the equilibrium expression for the reaction and determine the equilibrium constant.
For the equilibrium constant I wrote Kc = Reactants/products = (PH3^4)/(P4)(H2^6)
After finding the equilibrium concentrations for the reactants, my answer is 0.68. I need to know if this is right?
Would the answer be different if it wasn't a 1L vessel? If it was a 2L vessel?
the Kc will be 0.683 dm^2 per mol when the reaction takes place in the 1litre vessel.
Kc=((0.2/1)^4 / (0.5/1)^5 * (0.15/1) =0.683 dm^2 per mol
If you change the volume of this reaction the Kc will change..
Kc =(0.2/2)^4 / (0.5/2)^5 * (0.15/2) =1.37 dm^2 per mol
