Question

temp is : 298K

QUESTION 9 Please show all work. For full credit provide equations used. Please use correct significant figures and correct units in answers. The equilibrium constant for the reaction, 2Fe3+ (aq) + Hg22+ (aq) = 2Fe2+ (aq) + 2Hg2+ is Kc = 9.1 x 10-6 @298K a. What is AGO at this temperature? b. If reactants and products in their standard state concentrations (1M) are mixed, in which direction does the reaction proceed? Provide a numerical justification for your response. C. Calculate AG when [Fe3+] = 0.20 M, [Hg22+] = 0.010 M, [Fe2+] = 0.010 M, and (Hg2+] = 0.025 M. In which direction will the reaction proceed to achieve equilibrium? Provide equation and values used in calculating AG.

Answer 1

Reaction:
2Fe³⁺(aq) + Hg₂²⁺ (aq) $\rightleftharpoons$ 2Fe²⁺ (aq) + 2Hg²⁺ (aq) K_c = 9.1 x 10^-6 at 298 K

a.

This is the standard temperature, so:
$\Delta G^{o} = -RTlnK$

Here, R = 8.314 J/mol-K , T = 298 K , K = 9.1 x 10^-6  

AG° = -8.314J/mol. K x 298K x In(9.1 x 10-6

AG° = -2477.57J/mol x [In(9.1) +In(10-6)

AG° = -2477.57J/mol x [2.2 – 6(in 10)

$\Delta G^{o} = - 2477.57 J/mol \times[ 2.2 -6 (2.3)]$

$\Delta G^{o} = - 2477.57 J/mol \times[ 2.2 -13.8]$

$\Delta G^{o} = - 2477.57 J/mol \times[-11.6]$

AG° = +28739.8J/mol

$\Delta G^{o} = +28. 7 kJ/mol$

b.

Q for the given reaction can be written as:

$Q = \frac{ [Fe^{2+}]^{2} [Hg^{2+}]^{2}}{[Fe^{3+}]^{2} [Hg_{2}^{2+}]}$

If all concentrations are 1 . Then, Q = 1 and as we know K_c = 9.1 x 10^-6 .

Hence Q > K -> so reaction will move towards left (backward direction) to produce more of the reactants.

c.

$\Delta G = \Delta G^{o} +RTlnQ$

Here $\Delta G^{o} = +28. 7 kJ/mol = 28700 J/mol$ , R = 8.314 J/mol.K , T = 298K  
and, Q can be calculated using given concentrations: [Fe³⁺] =0.20M [Hg₂²⁺] =0.010M [Fe²⁺] = 0.010M [Hg²⁺] = 0.025M

$Q = \frac{ [Fe^{2+}]^{2} [Hg^{2+}]^{2}}{[Fe^{3+}]^{2} [Hg_{2}^{2+}]} = \frac{(0.010)^{2} (0.025)^{2}}{(0.20)^{2}(0.010)}$

$Q = \frac{10^{-4} \times 6.25 \times 10^{-4}}{4 \times 10^{-2}\times 10^{-2} }$

$Q = 1.56 \times 10^{-4}$

Here, Q > K -> so reaction will move towards left (backward direction) to produce more of the reactants.

Now,

$\Delta G = \Delta G^{o} +RTlnQ$

$\Delta G = 28700J/mol + 8.314 J/mol.K \times 298K \times ln(1.56 \times 10^{-4})$

$\Delta G = 28700J/mol + 2477.6 J/mol \times[ ln(1.56) + ln(10^{-4})]$

$\Delta G = 28700J/mol + 2477.6 J/mol \times[ 0.44 -4 \times ln(10)]$

$\Delta G = 28700 + 2477.6 \times[ 0.44 -4 \times 2.3]J/mol$

$\Delta G = 28700 + 2477.6 \times[ 0.44 -9.2]J/mol$

$\Delta G = 28700 + 2477.6 \times[-8.76]J/mol$

$\Delta G = 28700 -21703.8 \, \, J/mol$

$\Delta G = 6996.2 \, \, J/mol$

$\Delta G = 6. 99 \, \, kJ/mol$