C p d p 2 a p b p p o c d 2 p a b p o k p p o since

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C p ( ) ! D p ( ) 2 ! A p ( ) ! B p ( ) p o = ! C ! D ( ) 2 p ! A ! B p o = K ! p p o Since the total pressure of the system at equilibrium is equal to standard pressure K p = K χ and therefore K p is 9.6. d) The standard Gibbs energy of this reaction can be determined using: ( ) ( ) ( ) -1 -1 -1 ln 8.314 J K mol 298 K ln 9.550 5.6 kJ mol o r m G RT K ! = " = " = " Therefore the standard Gibbs energy of this reaction is -5.6 kJ mol -1 . The fact that it is negative is consistent with our knowledge that at equilibrium you will have more products than reactants. 3) If, in addition to the other gases described in the initial state for the system in question 2, there is also 0.01 mol of gas C, what would be the Gibbs energy for this reaction before it is allowed proceed (i.e. under initial conditions)? Since this is a non-equilibrium condition we must use: ! r G m = ! r G m o + RT ln Q The reaction quotient equation is the same as the equilibrium constant equation determined above: Q = ! C ! D ( ) 2 p ! A ! B p o
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Mole fractions can be calculated by taking the initial number of moles (from first row of ICE table) and dividing by the total number of moles of gas initially present. n total = n A + n B + n C + n D = 2.00 + 1.00 + 0.01 + 3.00 ( ) mol = 6.01 mol ! A = 2.00mol 6.01 mol = 0.333 ! B = 1.00mol 6.01 mol = 0.166 ! C = 0.01 mol 6.01 mol = 0.0017 ! D = 3.00mol 6.01 mol = 0.499 These values can be substituted into the reaction quotient equation above (also remembering that the total pressure of the system is equal to standard pressure): Q = 0.0017 ( ) 0.499 ( ) 2 0.333 ( ) 0.166 ( ) = 7.66 ! 10 " 3 Put this along with the standard Gibbs energy determined in d) into the equation for Gibbs energy: ! r G m = " 5.6 kJ mol -1 + 10 " 3 kJ J -1 ( ) 8.314 J K -1 mol -1 ( ) 298 K ( ) ln 7.66 # 10 " 3 ( ) = " 17.6 kJ mol -1 The large, negative Δ G indicates that the reaction will go forward in the direction that it’s written under these initial conditions.
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