The formation constant for the following reaction is Kf = 5.6 108 at 25°C. Ni2+(aq) + 6 NH3(aq) Ni(NH3)62+(aq) (a) Determine G0 at this temperature. kJ/mol (b) If standard-state concentrations of the reactants and products are mixed, in which direction does the reaction proceed? The reaction proceeds to the . (c) Determine G when [Ni(NH3)62+] = 0.017 M, [Ni2+] = 0.0015 M, and [NH3] = 0.0058 M. kJ/mol (d) If these concentrations are mixed, in which direction will the reaction proceed to achieve equilibrium? The reaction proceeds to the .

a. ΔG⁰ = -49.9 kJ/mol

b. The reaction proceeds towards right direction

c. ΔG = 49.5 kJ/mol

d. The reaction proceeds towards left direction

Explanation:

a. It is possible to obtain ΔG⁰ from K reaction using:

ΔG⁰ = -RT ln K

Where R is gas constant 8.314 J/molK, T is absolute temperature (273.15 +25 = 298.15K)

Replacing:

ΔG⁰ = -8.314 J/molK 298.15K ln 5.6x10⁸

ΔG⁰ = -49.9 kJ/mol

b. Reaction quotient Q, defines concentrations of substances involved in an equilibrium, when Q = Kf, the reaction is in equilibrium, when Q < Kf the reaction moves towards right direction and vice versa.

At standard-state concentrations, Q = 1, as:

Q < Kf,  the reaction proceeds towards right direction

c. ΔG = ΔG⁰ + RT ln Q

Q = [Ni(NH₃)₆²⁺] / [NH₃]⁶[Ni²⁺]

Q = 0.017M / 0.0015M⁶×0.0058M

Q = 2.6x10¹⁷

ΔG = -49.9 kJ/mol + 8.314x10⁻³ kJ/molK 298.15K ln 2.6x10¹⁷

ΔG = 49.5 kJ/mol

d. As Q > Kf, the reaction proceeds towards left direction

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explanation:

each principal energy level above the first contains one s orbital and three p orbitals. a set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. therefore, the second level can contain a maximum of eight electrons - that is, two in the s orbital and 6 in the three p orbitals.

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explanation: