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For reversible reaction A(g)⇌B(g)+C(g),KP=0.111 atm at 1000K. If change in enthalpy (ΔH)for the reaction is 9.212 k.cal, then equilibrium constant Kpat 1200K is (Consider ΔH is independent of temperature)

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Detailed Solution

Let Kp1 and Kp2 are equilibrium constants at 1000 K and 1200 K respectively. logK2K1=ΔH2.303R1T1-1T2logKP20.111=9.212×10002.303×211000-11200logKP20.111=13KP20.111=1013KP2=2.134×0.111=0.237

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For reversible reaction A(g)⇌B(g)+C(g),KP=0.111 atm at 1000K. If change in enthalpy (ΔH)for the reaction is 9.212 k.cal, then equilibrium constant Kpat 1200K is (Consider ΔH is independent of temperature)