The standard enthalpy and entropy changes for the reaction in equilibrium for the forward reaction are given below.CO(g)+H2O(g)⇌CO2(g)+H2(g)ΔH300K∘=−41.16kJmol−1ΔS300K∘=−4.24×10−2kJmol−1ΔH1200K∘=−32.93kJmol−1Then, the incorrect statement is

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The reaction proceeds in the forward direction at 300 K

At 1200 K, reaction proceeds in the reverse direction

At 1200 K, Kp > I

At 300 K, the products will be favored more than reactants at equilibrium

answer is C.

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Detailed Solution

At 300K,ΔG∘=−41.16−300×−4.24×10−2<0⇒ spontaneous At 1200K,ΔG∘=−32.93−1200×−4.24×10−2>0⇒ Nonspontaneous

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