The formation constant for the following reaction is K_f = 5.6 x 10^8 at 25^degr

Question

The formation constant for the following reaction is K_f = 5.6 x 10^8 at 25^degree C. Ni^2 +(a q) + 6 NH_3(a q) rightleftarrows Ni(NH_3)_6^2+(aq) Determine AG^ degree at this temperature. kJ/mol If standard-state concentrations of the reactants and products are mixed, in which direction does the reaction proceed? The reaction proceeds to the Determine Delta G when [ Ni ( NH_ 3 )_ 6^2+] = 0.014 M, [Ni^2+] = 0.0011 M, and [NH_3] = 0.0052  M.  kJ/mol If these concentrations are mixed, in which direction will the reaction proceed to achieve equilibrium? The reaction proceeds to the

Explanation / Answer

a) delta G0 = -R*T*lnKf = -8.314*298*ln(5.6*108) = -49906.8 J/mole = -49.906 kJ/mole

b) At standard states, the product side will be favoured i.e. reaction will move in the forward direction.

c) Kq = [Ni(NH3)62+]/{[Ni2+]*[NH3]6} = 6.44*1014

Thus, delta G = delta G0 + R*T*lnKq = -49906 + 8.314*298*ln(6.44*1014) = 34575 J = 34.575 kJ

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