7.6. For the following equilibrium, K =6.3 x 1014 at 1000 K.
NO (g) + O3 —–> NO2 (g) + O2(g).
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc for the reverse reaction?
7.6. For the following equilibrium, K =6.3 x 1014 at 1000 K.
NO (g) + O3 —–> NO2 (g) + O2(g).
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc for the reverse reaction?
For reverse reaction, KC (reverse) = 1/ KC = 1 / 6.3 x 1014 = 1.59 x 10-15
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0.01 M NaOH,
M = 1 * 10-2

pOH = 2
pH = 2
Kp = Kc (RT)Dng
36 * 10–2 = Kc (0.0821 * 300)–1
Kc = 0.36 * 0.0821 * 300 = 8.86 » 9
A(g) ->B(g) + (g)
Initial moles n 0 &nbs
On increasing pressure, equilibrium moves in that direction where number of gaseous moles decreases.
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Chemistry Ncert Solutions Class 11th 2023
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