Solved Problem
| Qp = | p(PCl3)p(Cl2) |
| p(PCl5) |
PCl5(g)
PCl3(g) + Cl2(g) Kp(673 K) = 11.5 Step 1: Enter given initial pressures.
Step 2: Assign changes with x = change in p(Cl2)
Step 3: Divide equilibrium pressures by 100 kPa.
p(Cl2) = p(PCl3) = x/100 p(PCl5) = 2.00 - x/100
Step 2: Assign changes with x = change in p(Cl2)
Step 3: Divide equilibrium pressures by 100 kPa.
p(Cl2) = p(PCl3) = x/100 p(PCl5) = 2.00 - x/100
| 11.5 = | (x/100)2 |
| 2.00 - x/100 |
| p(PCl5)/kPa | p(Cl2)/kPa | p(PCl3)/kPa | |
| initial | 200 | 0 | 0 |
| change | -x | +x | +x |
| final | 200-x | x | x |
Step 7: Solve for x.
| x = | -1150 ± √1150² - 4 × (-230,000) |
| 2 |
Step 6: Cross-multiply gives:
23 x 104 - 11.5x × 102 = x2
x2 + 1150x - 230,000 = 0
23 x 104 - 11.5x × 102 = x2
x2 + 1150x - 230,000 = 0
Step 5: Cross multiply gives:
| 23 - | 11.5x | = | x2 |
| 100 | 1002 |
x = 174 (Discarding the negative values)
Therefore: p(Cl2) = (PCl3) = 174 kPa and p(PCl5) = 26 kPa
The tough maths is much less important than the chemistry before the solution of the quadratic!!!