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प्रश्न
For the reaction
\[\ce{SrCO3(s) <=> SrO(s) + CO2(g)}\]
the value of equilibrium constant Kp = 2.2 × 10-4 at 1002 K. Calculate Kc for the reaction.
उत्तर
For the reaction,
\[\ce{SrCO3(s) <=> SrO(s) + CO2(g)}\]
∆ng = 1 – 0 = 1
∴ Kp = Kc (RT)
2.2 × 10-4 = Kc (0.0821) (1002)
`"K"_"C" = (2.2 xx 10^-4)/(0.0821 xx 1002) = 2.674 xx 10^-6`
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