Question

Determine ΔG when [Ni(NH3)62+] = 0.010 M, [Ni2+] = 0.0010 M, and [NH3] = 0.0050 M. In which direction will the reaction proceed to achieve equilibrium?Ni2+(aq) + 6 NH3(aq) ⇌ Ni(NH3)6 2+(aq) is Kf = 5.6 × 108 at 25°

Answer 1

Given:

`Kf = 5.6  X  10^(8)`

The reaction is

`Ni^(+2) (aq) + 6NH_(3)(aq) --->  Ni(NH_(3) )_(6) ^(+2) (aq)`

We know that

Δ
`G^(0) = -RT ln K_(f)`

Δ
`G^(0) = -(8.314) (298) ln (5.6  X  10^(8))`

Δ
`G^(0) = - (298)(8.314)( 20.14) = -49906.8 J mol^(-1)`

For the given reaction the reaction quotient will be :

Q = \frac{[Ni(NH_{3} )_{6} ^{+2}] }{[Ni^{+2} ][NH_{3}]^{6}

On putting values

Q will be equal to =
`6.4 X 10^(14)`

The relation between standard and other free energy change is

ΔG = Δ
`G^(0) + RT lnQ`

Thus

ΔG = (-49906.8) + (8.314)(298) ( ln (
`6.4 X 10^(14)`)

ΔG = (-49906.8) + 84466

ΔG = 34535 J / mol K

ΔG is positive so reaction is not spontaneous in forward direction

It will proceed in reverse direction, it will move from product to reactant