Effective Equilibrium Constant
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N2+3H2⇌2NH3 K1
N2+O2⇌2NO K2
H2+12O2→H2O K3
The equilibrium constant (K) of the reaction:
2NH3+52O2K⇌2NO+3H2O, will be
- K1K33/K2
- K2K33/K1
- K2K3/K1
- K32K3/K1
An aqueous solution contains and . If the equilibrium constants for the formation of ions from is and that of from ions is then the concentration of ions in an aqueous solution is :
S(s)+O2(g)⇌SO2(g);K1=1052
2S(s)+3O2(g)⇌2SO3(g);K2=10129
Find the equilibrium constant for
2SO2+O2(g)⇌2SO3(g);K3
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JEE MAIN 2019
- 10154
- 10181
- 1025
- 1077
- 2SO3(g)⇌2SO2(g)+O2(g)
- N2(g)+3H2(g)⇌2NH3(g)
- PCl5(g)⇌PCl3(g)+Cl2(g)
- H2(g)+I2(g)⇌2HI(g)
- 70
- 80
- 90
- 100
- Forward direction
- Backward direction
- The reaction will always be in equilibrium
- none of these
Calculate KC for the given change at 600oC.
- 675.23 mol−1L
- 765.32 mol−1L
- 833.33 mol−1L
- 801.23 mol−1L
- 70
- 80
- 90
- 100
If we take 1 mole of each of the four gases in a 10 litre container, what would be equilibrium concentration of A2(g).
- \N
- 0.13
- 0.25
- 1
- 1(K1K2)
- 1(2K1K2)
- 1(4K1K2)
- (1K1K2)1/2
2H2S(g)⇌2H2(g)+S2(g)
is 0.0118 at 1300 K while the heat of dissociation is 597.4 kJ. The standard equilibrium constant of the reaction at 1200 K is:
- 10−4
- 10−6
- 10−8
- 10−10
If the equilibrium constant of the reaction 2HI ⇌ H2 + I2 is 0.25, then the
equilibrium constant of the reaction H2 + I2 ⇌ 2HI would be
1.0
2.0
3.0
4.0
At certain temperature, a compound AB2 dissociates as: 2AB2(g)→2AB(g)+B2(g) with a degree of dissociation α, which is very small compared to unity. The expression of Kp in terms of α and total pressure P is ?
α2P2
α2P3
α3P3
α3P2
2NH3(g)⇌N2(g)+3H2(g).
Then, the equilibrium constant, Kp is expressed as:
(Where pN2 = partial pressure of N2 and pNH3 is the partial pressure of NH3 )
- Kp=33×p2N2pNH3
- Kp=33×pN2p3H2p2NH3
- Kp=33×p4N2p2NH3
- Kp=3(3/2)×p4N2p2NH3
- Forward direction
- Backward direction
- The reaction will always be in equilibrium
- none of these
- 5×10−19
- 5×10−8
- 3×10−20
- 6×10−21
- logKp2Kp1=△H2.303R[T1−T2T1T2]
- logKp2Kp1=△H2.303R[T2−T1T1T2]
- logKp2Kp1=△H2.303R[T1T2T1−T2]
- logKp2Kp1=△H2.303R[T1T2T2−T1]
- 32
- 23
- 1
- None of these
- Neither KP nor α changes
- Both KP and α changes
- KP changes, but α does not change
- KP does not change, but α changes
A plot of In K against 1T (abscissa) is expected to be a straight line with intercept on ordinate axis equal to?
ΔS∘2.303 R
ΔS∘R
−ΔS∘R
R×ΔS∘
N2+3H2⇌2NH3; K1
N2+O2⇌2NO; K2
H2+12O2⇌H2O; K3
The equilibrium constant (K) of the reaction:
2NH3+52O2K⇌2NO+3H2O will be
- K2K33/K1
- K2K3/K1
- K32K3/K1
- K1K33/K2
- 25 and 50 min
- 50 and 100 min
- 70 and 35 min
- 150 and 300 min
Equilibrium constants K1 and K2 for the following equilibria are
NO(g) + 0.5O2(g) →K1 NO2(g) and
2NO2(g) →K2 2NO(g) + O2(g)
K2 = 1K1
K2 = K21
K2 = K12
K2 = 1K21
Consider the following equilibria at 25∘C, 2NO(g)⇌N2(g)+O2(g);K1=4×1030and NO(g)+12Br2(g)⇌NOBr(g);K2=1.4 mol−12L12. The value of K_C for the reaction (at same temperature) 12N2(g)+12O2(g)+12Br2(g)⇌NOBr(g) is :
3.5×10−31
2.8×1015
7.0×10−16
5.6×1030
- √RT
- RT
- 1RT
- 1.0
PtCl2−4+H2O⇌Pt(H2O)Cl−3+Cl−
At 25°C, it is found –d[PtCl2−4]dt=(3.9×10−5s−1))[PtCl2−4]−(2.1×10−3Lmol−1s−1)[Pt(H2O)Cl−3][Cl−]
Value of KC when fourth Cl− is complexed in the reaction is:
- 1.85×10−2
- 1.86×2
- 53.85
- 8.19×10−8