Relation between Kp and Kc
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Q. For a given exothermic reaction, Kp and K′p are the equilibrium constants at temperatures T1 and T2 respectively. Assuming that heat of reaction is constant in temperatures range between T1 and T2. It is readily observation that :
(Given, T2>T1)
(Given, T2>T1)
- KP>K′P
- KP=K′P
- K=1K′P
- KP<K′P
Q. A solution has a 1:4 mole ratio of pentane to hexane. The vapour pressures of the pure hydrocarbons at 200 C are 440 mm Hg for pentane and 120 mm Hg for hexane. The mole fraction of pentane in the vapour phase would be:
- 0.200
- 0.549
- 0.786
- 0.478
Q. The pressure necessary to obtain 50% dissociation of PCl5 at 400 K is numerically equal to how many times the value of the equilibrium constant Kp
Q.
At 450 K, Kp=2.0×1010/bar for the given reaction at equilibrium.
2SO2(g)+O2(g)↔2SO3(g)
What is Kc at this temperature?
Q.
For which of the following KP is less than Kc?
N2O4 2NO2
N2 + 3H2 2NH3
H2 + I2 2HI
CO + H2O CO2 + H2