PCl₅ ⇌ PCl₃ + Cl₂
K c = 1.844
3.0 moles of PCl₅ is introduced in a 1 L closed reaction vessel at 380 K. The number of moles of PCl₅ at equilibrium is ______________ × 10⁻³. (Round off to the Nearest Integer)

Solution

Chemistry ma’am · 4 steps

Let x mol of PCl5 break up, write Kc in terms of x, and solve the quadratic for x.

1. Set up the table. Let x mol of PCl₅ dissociate. At equilibrium you have (3 - x) mol PCl₅, x mol PCl₃ and x mol Cl₂. The vessel is 1 L, so each number of moles is also the concentration in mol L⁻¹.

   K_c = [PCl₃][Cl₂]/[PCl₅] = x²/(3 - x) = 1.844

2. Form the quadratic. Cross-multiplying gives x² = 5.532 - 1.844x, that is x² + 1.844x - 5.532 = 0.
3. Solve for x. x = (-1.844 + √(1.844² + 4 × 5.532))/2 = (-1.844 + √25.528)/2 = (-1.844 + 5.0526)/2 = 1.6043. Keep only the positive root, because a negative amount of PCl₃ is not possible.
4. So. Moles of PCl₅ left = 3 - 1.6043 = 1.3957 mol = 1395.7 × 10⁻³ mol, which rounds to 1396.

The common slip: 1604. Reporting x, the moles of PCl₅ that dissociated, gives 1604. The question asks for what is left, which is 3 - x.

The idea to keep. For dissociation problems, let x be the amount that reacts and write every species in terms of x. Then check what the question asks for: the amount that reacted or the amount left.

Answer from the NTA final answer key. Solution by PrepWiser.