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For the reaction CH4(g) + H2O(g)3H2(g) + CO(g) H° = 206.1 kJ and S° = 214.7...

For the reaction

CH4(g) + H2O(g)3H2(g) + CO(g)

H° = 206.1 kJ and S° = 214.7 J/K The equilibrium constant for this reaction at 264.0 K is

Assume that H° and S° are independent of temperature.

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Answer #1

Als) we know that AG - AM-TASO where AG= Standard Gibbs free energy change AM = Standard enthalpy change T- Temperature to k= -149,419.2 2194.896 =-68-0757539 . In K = -68.0757539 (2x = 2:303loga) 2.303 log K = -68-07575 39 logka –68.0757539 2.303 =

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