The equilibrium constant for the reaction,

Question

The equilibrium constant for the reaction,

2 Fe3 (aq) Hg22 (aq) <=> 2 Fe2 (aq) 2 Hg2 (aq)

is 9.1 x 10-6(aq) at 298 K. Calculate Delta G in kJ when

[Fe3 (aq)] = 0.481

[Hg22 (aq)] = 0.0454

[Fe2 (aq)] = 0.0485

[Hg2 (aq)] = 0.0534

Answer 1

First, you need to find delta G at standard conditions with the equation

delta G at standard = -RTlnK

also, delta G = delta G at standard + RTlnQ

Substituting equation 1 into equation 2 yields

delta G = -RTlnK + RTlnQ

delta G = RTlnQ - RTlnK

delta G = RT(lnQ-lnK)

delta G = RTln(Q/K)

Q is calculated by using the provided concentrations:

Q = ([Hg2]^2*[Fe2]^2)/([Hg22]*[Fe3]^2)

Plug the concentrations into the Q expression, then use the equation delta G = RTln(Q/K) to find delta G (R is the gas law constant, 8.31 J/mol*K).

Also, the answer will be in Joules, so make sure to convert it to kJ by dividing by 1000.  

Answer 2

2Fe3+  +  Hg2 2+  &lt;==&gt;  Fe2+  +  2Hg2+

ΔG = -RT ln(K)

ΔG = - 8.314J/K mol x 298K x ln(K)

ΔG = 32000J  or  32kJ

Answer 3

Actually

i am not sure

32000J or 32kJ

Answer 4

Delta g= Delta g_0 +RTln(Q)

=-RTln(K)+RTlnQ

Q is just [Hg22 (aq)]^2*[Fe2 (aq)]^2/([Fe3 (aq)]^2*[Hg2 (aq)])=6.3858E-4

so plug this all in:

10.532kj