Chemical EquilibriumHard
Question
For a gaseous equilibrium 2A(g) $\rightleftharpoons$2B(g) + C(g), KP has a value 1.8 at 700 K. The value of KC for the equilibrium 2B(g) + C(g) $\rightleftharpoons$2A(g) at that temperature is about
Options
A.0.031
B.32
C.57.4
D.103.3
Solution
KP = KC.(RT)ΔHg
or 1.8 = KC × (0.0821 × 700)′ ⇒ KC = 0.031
Hence, for reverse reaction, $K_{C}' = \frac{1}{K_{C}} = 31.93$
Application of Equilibrium Constant
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