Calculate the equilibrium concentrations of reactant and products when 0.331 moles of PCl3 and 0.331 moles of Cl2 are introduced into a 1.00 L vessel at 500 K?
The equilibrium constant, Kc, for the following reaction is 83.3 at 500 K.
PCl3(g) + Cl2(g) PCl5(g)
Calculate the equilibrium concentrations of reactant and products when 0.331 moles of PCl3 and 0.331 moles of Cl2 are introduced into a 1.00 L vessel at 500 K.
PCl3=?M
Cl2=?M
PCl5=?M
The equilibrium constant, Kc, for the following reaction is 83.3 at 500 K.
PCl3(g) + Cl2(g) PCl5(g)
Calculate the equilibrium concentrations of reactant and products when 0.331 moles of PCl3 and 0.331 moles of Cl2 are introduced into a 1.00 L vessel at 500 K.
PCl3=?M
Cl2=?M
PCl5=?M
1 Answer
#["PCl"_3]_(eq) = ["Cl"_2]_(eq) = "0.0573 M"#
#["PCl"_5]_(eq) = "0.2737 M"#
Construct the ICE table:
#"PCl"_3(g) + "Cl"_2(g) " "rightleftharpoons" " "PCl"_5(g)#
#"I"" "0.331" "" "0.331" "" "" "" "0#
#"C"" "-x" "" "-x" "" "" "" "+x#
#"E"" "0.331-x" "0.331-x" "" "x#
So then, the mass action expression becomes:
#K_P = x/(0.331-x)^2 = 83.3#
#x = 83.3(0.109561 - 0.662x + x^2)#
#0 = 83.3x^2 - 56.1446x + 9.12643#
Solve the quadratic. You should get
#x = "0.4003 M", "0.2737 M"#
Which one doesn't make physical sense? (i.e. which one gives you NEGATIVE concentration?)
So you should get
#["PCl"_3]_(eq) = ["Cl"_2]_(eq) = "0.0573 M"#
#["PCl"_5]_(eq) = "0.2737 M"#
