%3Cp%3E%3Cspan%20class%3D%22c1%22%3E1.%20At%20a%20certain%20temperature%2C%200.9
ID: 822113 • Letter: #
Question
%3Cp%3E%3Cspan%20class%3D%22c1%22%3E1.%20At%20a%20certain%20temperature%2C%200.920%20mol%20of%20SO3%0Ais%20placed%20in%20a%204.50-L%20container.%3C%2Fspan%3E%3C%2Fp%3E%0A%3Cp%3E%3Cspan%20class%3D%22c1%22%3E%3Cbr%20%2F%3E%3C%2Fspan%3E%3C%2Fp%3E%0A%3Cp%3E%3Cspan%20class%3D%22c1%22%3E2SO3(g)%20---%26gt%3B%202SO2(g)%20%2B%20O2%20(g)%0A%20%3C%2Fspan%3E%3Cspan%20class%3D%22c1%22%3EAt%20equilibrium%2C%200.140%20mol%20of%20O2%20is%0Apresent.%20Calculate%20Kc.%20%3D%3C%2Fspan%3E%3C%2Fp%3E%0A%3Cp%3E%3Cspan%20class%3D%22c1%22%3E____%20%3F%3C%2Fspan%3E%3C%2Fp%3E%0A%3Cp%3E%3Cspan%20class%3D%22c1%22%3E%3Cbr%20%2F%3E%3C%2Fspan%3E%3C%2Fp%3E%0A%3Cp%3E%3Cspan%20class%3D%22c1%22%3E2.%26nbsp%3B%3C%2Fspan%3E%3Cspan%20class%3D%22c1%22%3EPhosphorus%0Apentachloride%20decomposes%20according%20to%20the%20chemical%0Aequation%26nbsp%3B%3C%2Fspan%3E%3C%2Fp%3E%0A%3Cp%3E%3Cspan%20class%3D%22c1%22%3EPCl5(g)%20----%26gt%3B%20PCl3(g)%20%2B%20Cl2(g)%20%26nbsp%3B%20%26nbsp%3B%0A%26nbsp%3BKc%20%3D%201.80%20at%20250%20C%3C%2Fspan%3E%3C%2Fp%3E%0A%3Cp%3E%3Cspan%20class%3D%22c1%22%3E%3Cbr%20%2F%3E%3C%2Fspan%3E%3C%2Fp%3E%0A%3Cp%3E%3Cspan%20class%3D%22c1%22%3EA%200.245%20mol%20sample%20of%20PCl5(g)%20is%20injected%20into%0Aan%20empty%202.95%20L%20reaction%20vessel%20held%20at%20250%20%C2%B0C.%20Calculate%20the%0Aconcentrations%20of%20PCl5(g)%20and%20PCl3(g)%20at%20equilibrium.%3C%2Fspan%3E%3C%2Fp%3E%0A%3Cp%3E%3Cspan%20class%3D%22c1%22%3E%3Cbr%20%2F%3E%3C%2Fspan%3E%3C%2Fp%3E%0A%3Cp%3E%3Cspan%20class%3D%22c1%22%3EPCl5%20%3D%20_____%20%26nbsp%3B%3F%20M%3C%2Fspan%3E%3C%2Fp%3E%0A%3Cp%3E%3Cspan%20class%3D%22c1%22%3E%3Cbr%20%2F%3E%3C%2Fspan%3E%3C%2Fp%3E%0A%3Cp%3E%3Cspan%20class%3D%22c1%22%3EPCl3%20%3D%20______%20%3F%20M%26nbsp%3B%3C%2Fspan%3E%3C%2Fp%3E%0A%3Cp%3E%3Cspan%20class%3D%22c1%22%3E%3Cbr%20%2F%3E%3C%2Fspan%3E%3C%2Fp%3E%0AExplanation / Answer
Reaction : 2SO3(g) ---> 2SO2(g) + O2 (g)
Initial : 0.920 0 0
Change : - 2x +2x +x
Final : 0.920 - 2x +2x +x
Now,
x = 0.140
2x = 0.280
0.920 - 2x = 0.640
Kc = [0.14 / 4.5] * [0.28/4.5]^2 / [0.64 / 4.5]^2
= 3.811 * 10^-3
Now,
Reaction : PCl5(g) ----> PCl3(g) + Cl2(g) Kc = 1.80 at 250 C
Initial : 0.245 0 0
Change : - x + x + x
Final : 0.245-x +x +x
Kc = [x/2.95] * [x /2.95] / { 0.245 - x / 2.95 } = 1.80
x^2 = 1.30095 - 5.31 x
x = 0.2346 moles
thus PCl5 = 0.245 .0- 0.234 = 0.011 moles = 3.728 * 10^-3 M
PCl3 = 0.234 moles = 0.079 M
Hope this helps. Please rate it ASAP. Thanks
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