equilibrium constant (Kp)
Kp
(PC)^c(PD)^d/(PA)^a(PB)^b
Calculate the Kp value and its units for the decomposition of PCl5 gas at 500K. The partial pressures of each gas are: pPCl5 = 188kPa pPCl3 = 263 kPa pCl2 = 263 kPa
PCl5 <> PCl3 + Cl2 Kp = (pPCl3) x (pCl2) / (pPCl5) Kp = (263) x (263) / 188 = 367.92 Kp = 368 units = kPa x kPa / kPa = kPa 368 kPa
when 2.00 moles of PCl5 is heated in a sealed container, the equilibrium mixture contains 1.25 moles of chlorine PCl5 <> PCl3 + Cl2 Calculate the mole fraction of PCl5
at equilibrium: Cl2 - 1.25 moles PCl3 - 1.25 moles PCl5 - 2-1.25 = 0.75 mole fraction = 0.75 / 1.25 + 1.25 + 0.75 = 0.2307 = 0.231 moles
SO2 + 1/2O2 <> SO3 ΔH = -197 kJ mol-1 Kp = (pSO3) / (pSO2) x (pO2)^1/2 how would changing concentration affect this equilibrium
changing concentration = changing pressure in a gaseous equilibrium no change - Kp is a constant so not affected by anything other than temperature
SO2 + 1/2O2 <> SO3 ΔH = -197 kJ mol-1 Kp = (pSO3) / (pSO2) x (pO2)^1/2 how would changing temperature affect this equilibrium
decrease temperature - equilibrium shifts forward (exothermic) so SO3 mole fraction increases so partial pressure increases so top portion of equation increases so Kp increases increase temperature - equilibrium shifts backwards (endothermic direction) so SO2 and O2 mole fractions increase so partial pressures increase so bottom portion of equation increases so Kp decreases
partial pressure of a gas
mole fraction of gas x total pressure of the mixture
A mixture of 8.02g of nitrogen gas and 2.42g of oxygen gas is contained inside a sealed vessel. Given that the total pressure of the mixture is 921 kPa, what is the partial pressure of the oxygen.
moles N2 = 8.02/28 = 0.2864 moles O2 = 2.42/32 = 0.075625 M.F. O2 = 0.075625/0.2864+0.075625 = 0.21 P.P. O2 = 0.21 x 921 = 192.365 kPa = 192 kPa
SO2 + 1/2O2 <> SO3 ΔH = -197 kJ mol-1 Kp = (pSO3) / (pSO2) x (pO2)^1/2 how would a catalyst affect the equilibrium of this reaction
no effect - makes reaction reach equilibrium faster but no effect on the position as it increases forwards and backwards reactions equally
mole fraction
number of moles of the gas / total moles of gas