Derivation of Kp and Kc
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N2(g)+3H2(g)⇌2NH3(g)
The equilibrium constant of the above reaction is Kp. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by: (Assume that ) PNH3<<Ptotal at equilibrium)
- 33/2K1/2pP224
- 33/2K1/2pP222
- K1/2pP222
- K1/2pP224
The dissociation equilibrium of a gas can be represented as:
The degree of dissociation is and is small compared to . The expression relating the degree of dissociation with equilibrium constant and total pressure is?
KMnO4 + FeSO4 + H2SO4 -----> Fe2(SO4)3 + K2SO4 + H2O
[Given : R=0.082 L atm K–1mol–1]
- 6.24
- 12.13
- 15.24
- 2.25
- 0.586
- 0.121
- 0.150
- 0.254
CaCO3(s)⇌CaO(s)+CO2(g)
The experiment was carried out in the temperature range 800−900oC, Equilibrium constant Kp follows the relation. log Kp=7.282−8500T
where, T is temperature in Kelvin. At what temperature the decomposition will give CO2(g) at 1 atm?
- 894.11oC
- 794.66oC
- 987.18oC
- 635.45oC
Write the expression for the equilibrium constant, KC for each of the following reactions:
(i) 2NOCl(g)↔2NO(g)+Cl2(g)
(ii) 2Cu(NO3)2(aq)↔2CuO(S)+4NO2(g)+O2(g)
(iii) CH3COOC2H5(aq)+H2O(l)↔CH3COOH(aq)+C2H5OH(aq)
(iv) Fe3+(aq)+3OH−(aq)↔Fe(OH)3(s)
(v) I2(s)+5F2↔2IF5
- PCl5(g)→PCl3(g)+Cl2(g)
- 2SO2(g)+O2(g)→2SO3(g)
- H2(g)+I2(g)⇌2HI(g)
- N2(g)+3H2(g)⇌2NH3(g)
- n−m
- m+n
- 2n−m
- 12m+n
2H2O(g)+2Cl2(g)⇌4HCl(g)+O2(g) is 0.03 atm at 427∘C, when the partial pressure are expressed in atmosphere, then the value of Kc for the same reaction is
- 5.23×10−4
- 7.34×10−4
- 3.2×10−3
- 5.43×10−5
Here,
t1/2 is the half life period of the reaction
- For a first order reaction, t1/2 is directly proportional to initial concentration
- For a zero order reaction , t1/2 is inversely proportional to initial concentration
- For a second order reaction, t1/2 is inversely proportional to initial concentration
- All of the above
Why do solids not affect equilibrium?
H2O (g)+CO (g)⇌H2 (g)+CO2 (g). What is the value of equilibrium constant (Kc) for the reaction?
- 44
- 2.22
- 4.4
- 0.44
- 0.147
- 0.026
- 0.05
- 0.073
At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl5is8.3×10−3. If decomposition is depicted as,
PCl5(g)↔PCl3(g)+Cl2(g) ΔrH∘=124.0kJmol−1
Write an expression for Kc for the reaction.
What is the value of Kc for the reverse reaction at the same termperature?
What would be the effect on Kc if
(i) more PCl5 is added
(ii) pressure is increased?
(iii) The temperature is increased?
2HI⇌H2+I2. The equilibrium constant for dissociation is -
- 0.282
- 0.0786
- 0.0199
- 1.99
Ammonium carbonate decomposes as
NH2COONH4 (s) ⇋ 2NH3 (g) + CO2 (g) For the reaction, KP=2.9×10−5 atm3. Calculate the total pressure at equilibrium if 1 mole of NH2COONH4is taken.
- 0.0766 atm
- 0.0582 atm
- 0.0388 atm
- 0.0194 atm
In the reversible reaction A + B ⇌ C + D, the concentration of each C and D at
equilibrium was 0.8 mole/litre, then the equilibrium constant KC will be
6.4
0.64
1.6
16.0
A sample of HI_{(g)} is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI_{(g)} is 0.04 atm. What is K_p for the given equilibrium?
2HI(g)↔H2(g)+I2(g)
Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:
2BrCl↔Br2(g)+Cl2(g)
Fro which Kc=32 at 500 K.
If initially pur BrCl is present at a concentration of 3.3×10−3mol−1, what is its molar concentration in the mixture at equilibrium?
2SO3(g)⇌2SO2(g)+O2(g)
is 1.8×10−3kPa. What is the numerical value of KC for this reaction at the same temperature?
R=0.0821 L atm mol−1 K−1
- 3.09×10−7 mole litre−1
- 9.03×10−7 mole litre−1
- 5.05×10−9 mole litre−1
- 5.05×10−5 mole litre−1
- 20 atm
- 24 atm
- 12 atm
- 16 atm
- 8 times
- 16 times
- 64 times
- 4 times.
CO2+C⇌2CO
Calculate the value of Kp if total pressure at equilibrium is 0.8 atm.
- 2.5 atm
- 1.1 atm
- 0.5 atm
- 1.8 atm
Kp=0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C2H6 when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?
C2H6(g)↔C2H4(g)+H2(g)
- 1
- RT
- (RT)1/2
- (RT)−1/2