PCl5 in a heated glass bulb
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a) A weighted amount of PCl5 is sealed in a 100.0mL glass bulb and heated to 250'C. When the system reaches equilibrium, the pressure in the bulb is 0.895atm. If Kp = 2.15 at 250'C, what is the partial pressure of each of the three gases at equilibrium?
PCL5(g) ===========> PCl3(g) + Cl2(g)
b) Calculate Kc at 250'C.
c) If delta H for the reaction is +87.9 kJ/ mol, calculate Kp at 200'C.
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Solution Summary
The Kc and Kp for the reaction of PCl5 are calculated.
Solution Preview
Let Px, Py and Pz be the partial pressures of PCl5, PCl3 and Cl2. For every 'x' mol of PCl3 that forms, 'x' mol of CL2 should also form. Hence Py =Pz
Total pressure = 0.895 atm. Hence Px + ...
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