Advertisements
Advertisements
प्रश्न
For the following equilibrium, Kc = 6.3 × 1014 at 1000 K
\[\ce{NO (g) + O3 (g) ⇌ NO2 (g) + O2 (g)}\]
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc, for the reverse reaction?
उत्तर
It is given that `"K"_"c"` for the forward reaction is `6.3 xx 10^(14)`
Then, `"K"_"c"` for the reverse reaction will be,
`"K'"_"C" = 1/"K"_"C"`
`= 1/(6.3 xx 10^(14))`
`= 1.59 xx 10^(-15)`
APPEARS IN
संबंधित प्रश्न
Describe the effect of addition of H2 on the equilibrium of the reaction:
\[\ce{2H2(g) + CO (g) ⇌ CH3OH (g)}\]
Describe the effect of Addition of CH3OH on the equilibrium of the reaction:
\[\ce{2H2(g) + CO (g) ⇌ CH3OH (g)}\]
Describe the effect of Removal of CO on the equilibrium of the reaction:
\[\ce{2H2(g) + CO (g) ⇌ CH3OH (g)}\]
Describe the effect of Removal of CH3OH on the equilibrium of the reaction:
\[\ce{2H2(g) + CO (g) ⇌ CH3OH (g)}\]
Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high-temperature steam. The first stage of two-stage reaction involves the formation of CO and H2. In the second stage, CO formed in the first stage is reacted with more steam in water gas shift reaction,
\[\ce{CO (g) + H2O (g) ⇌ CO2 (g) + H2 (g)}\]
If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam such that `"p"_"CO" = "p"_("H"_2"O")` = 4.0 bar, what will be the partial pressure of H2 at equilibrium? Kp = 10.1 at 400°C.
Which of the following is NOT a general characteristic of equilibria involving physical processes?
