Given the reaction: PCl5 (g) -> PCl3 (g) + Cl2 (g) a) Calculate Kp for this reac
ID: 526206 • Letter: G
Question
Given the reaction: PCl5 (g) -> PCl3 (g) + Cl2 (g)
a) Calculate Kp for this reaction at 298 K, and at 373 K
b)Calculate delta Go for this reaction at 373 K
c) PCl5 gas, PCl3 gas, and Cl2 gas are added to a flask at 373 K. Their partial pressures are 0.670 bar, 0.320 bar, and 0.450 bar, respectively. Calculate delta G for this reaction at the instant the gases begin to react.
d Calculate the equilibrium partial pressure of each gas at 373 K for part c)
PCl5 (g) delta Gfo (kJ/mol) at 298 K (-305.0), delta Hfo (kJ/mol) at 298 K (-374.9)
PCl3 (g) delta Gfo (kJ/mol) at 298 K (-267.8), delta Hfo (kJ/mol) at 298 K (-287.0)
Cl2 (g) delta Gfo (kJ/mol) at 298 K (0), delta Hfo (kJ/mol) at 298 K (0)
Explanation / Answer
(a)
deltaG0rxn = deltaG0f(PCl3) + deltaG0f(Cl2) - deltaG0f(PCl5)
= - 267.8 - 0 - ( - 305.0 )
= + 37.2 kJ
deltaG0 = - R T lnK
37.2 = - 0.008314 * 298 * lnK
lnK = - 15.0
K = e-15.0
K = 3.06 * 10-7
deltaH0 = deltaH0f(PCl3) + deltaH0f(Cl2) - deltaH0f(PCl5)
= - 287.0 - 0 - ( - 374.9 )
= + 87.9 kJ
(b) According ot van't Hoff's equation,
lnK2/K1 = (deltaH0/R)*[(1/T1) - (1/T2)]
lnK2/(3.06*10-7) = (87.9/0.008314) * [(1/298) - (1/373)]
K2 / (3.06*10-7) = e1.73
K2 = 5.64 * 3.06 * 10-7
K2 = equilibrium constant at 373 K = 1.73 * 10-6
(c)
Q = PCl3 * PCl2 / PCl5
Q = 0.320 * 0.450 / 0.670
Q = 0.215
deltaG = deltaG0 + R T lnK
deltaG = 37.2 + 0.008314 * 373 * ln(1.73 * 10-6)
deltaG = - 3.94 kJ
(d)
PCl5 (g) = PCl3 (g) + Cl2 (g)
0.670 0.320 0.450
0.670-x x x
Kp = PPCl3 * PCl2 / PPCl5
1.73 * 10-6 = x * x / (0.670-x)
1.73 * 10-6 = x2 / 0.670 (since Kp is small, 0.670 -x is taken as 0.670)
x = 1.08 * 10-3 bar
At euilibrium,
x = PPCl3 = PCl2 = 1.08 * 10-3 bar
PPCl5 = 0.670 - 0.00108 = 0.669 bar
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