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The question is incomplete, so I tried to find a similar problem online. It is shown in the attached picture. The reaction is
PCl₃ + Cl₂ ⇆ PCl₅
Then, we use the ICE (Initial-Change-Excess) approach as follows:
PCl₃ + Cl₂ ⇆ PCl₅
I 0.5 0.5 0.3
C -x -x +x
E 0.5-x 0.5-x 0.3+x
Total pressure: 0.5 - x + 0.5 - x + 0.3 + x = 1.3
Kp = [PCl₅]/[PCl₃][Cl₂]
0.18 = (0.3+x)/(0.5-x)²
Solving for x,
x= 0.21
Partial pressures would be:
<em>PCl₃ = 0.5 - 0.21 = 0.29 atm</em>
<em>Cl₂ = 0.5 - 0.21 = 0.29 atm</em>
<em>PCl₅ = 0.3+0.21 = 0.51 atm</em>
PCl₃ + Cl₂ ⇆ PCl₅
Then, we use the ICE (Initial-Change-Excess) approach as follows:
PCl₃ + Cl₂ ⇆ PCl₅
I 0.5 0.5 0.3
C -x -x +x
E 0.5-x 0.5-x 0.3+x
Total pressure: 0.5 - x + 0.5 - x + 0.3 + x = 1.3
Kp = [PCl₅]/[PCl₃][Cl₂]
0.18 = (0.3+x)/(0.5-x)²
Solving for x,
x= 0.21
Partial pressures would be:
<em>PCl₃ = 0.5 - 0.21 = 0.29 atm</em>
<em>Cl₂ = 0.5 - 0.21 = 0.29 atm</em>
<em>PCl₅ = 0.3+0.21 = 0.51 atm</em>