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