Given the following values for components of a system, at 410 K:
For PCl5, ΔH°f = -370.0 kJ mol−1, S° = 375.0 J mol−1 K−1.
For PCl3, ΔH°f = -293.0 kJ mol−1, S° = 320.5 J mol−1 K−1.
For Cl2, ΔH°f = 000.0 kJ mol−1, S° = 230.0 J mol−1 K−1.
Calculate a value for Kp at 410 K for PCl5(g)
PCl3(g)+ Cl2(g)Hint: Find ΔG, then Kp.
Correct Answer:
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