For the reaction PCl3(g) + Cl2(g) PCl5(g) at a particular temperature, Kc = 24.3. ?
For the reaction PCl3(g) + Cl2(g) PCl5(g) at a particular temperature, Kc = 24.3. Suppose a system at that temperature is prepared with [PCl3] = 0.10 M, [Cl2] = 0.15 M, and [PCl5] = 0.60 M. What are the equilibrium concentrations of PCl3(g), Cl2(g) and PCl5(g)?
- Roger the MoleLv 71 month agoFavorite Answer
PCl3(g) + Cl2(g) → PCl5(g) [balanced as written]
Kc = [PCl5] / ([PCl3] [Cl2])
Let "z" be the concentration (in M) of PCl5 lost in the reaction.
Then the concentrations at equilibrium are:
[PCl5] = 0.60 - z
[PCl3] = 0.10 + z
[Cl2] = 0.15 + z
Substitute these expressions into the formula for Kc:
24.3 = (0.60 - z) / ((0.10 + z) x (0.15 + z))
Solve for z algebrically, and discard the negative solution:
z = 0.0301617 = 0.03 M
[PCl5] = 0.60 - 0.03 = 0.57 M
[PCl3] = 0.10 + 0.03 = 0.13 M
[Cl2] = 0.15 + 0.03 = 0.18 M