Part A
The change in enthalpy (ΔrHo) for a reaction is -32 kJmol−1 .
The equilibrium constant...
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Chemistry
Part A
The change in enthalpy (ΔrHo) for a reaction is -32 kJmol−1 .The equilibrium constant for the reaction is 4.6×103 at 298 K. Whatis the equilibrium constant for this reaction at 661 K?
Part B
Consider the following reaction: CH3OH(g)⇌CO(g)+2H2(g)
Calculate ΔrG for this reaction at 125 ∘C under the followingconditions: PCH3OH= 0.890 bar PCO= 0.145 bar PH2= 0.195 bar
Answer & Explanation
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3.7 Ratings (457 Votes)
A We have relation ln K2K1 dH R 1T1 1T2 T1 298 K T2 661 K dHo 32 KJmol 32000
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