During chemical weathering, forsterite is dissolved by the carbonic acid in rain

Question

During chemical weathering, forsterite is dissolved by the carbonic acid in rainwater. The weathering reaction is:

Mg2SiO4 forsterite + 4H2CO3 (aq) ? 2Mg2+ + 4HCO3- + H4SiO4 (aq) Use the thermodynamic data from Appendix II, source 2, for the following calculations.

4a. Calculate the Keq for this weathering reaction at 25 C.

4b. If the reaction is at equilibrium, using Le Cha?teliers principle, predict what

would happen if Mg2+ ions were added to the solution.

4c. Using Le Cha?teliers principle, predict what would happen to the equilibrium

constant if the reaction occurred at a higher temperature.

4d. Calculate the Keq for this reaction at 40 C. Does the solubility of forsterite

increase or decrease with increasing temperature? How does this result compare with your prediction in part (c).

Appendix II

standard state = 298.15 k 10^5 pa

Mg2SiO4 =

Delta G f= 2056.7 kJ mol^-1

Delta H f= 2175.7 kJ mil ^-1

S= 95.2 J mol^-1 k^-1

4H2CO3 =

Delta G f= 623.14 kJ mol^-1

Delta H f= 699.09 kJ mil ^-1

S= 189.31 J mol^-1 k^-1

2mg^2+ =

Delta G f= 455.4 kJ mol^-1

Delta H f= 467.0 kJ mil ^-1

S= 137 J mol^-1 k^-1

4HcO3^- =

Delta G f= 586.8 kJ mol^-1

Delta H f= 689.93 kJ mil ^-1

S= 98.4 J mol^-1 k^-1

H4SiO4 =

Delta G f= 1307.9 kJ mol^-1

Delta H f= 1457.3 kJ mil ^-1

S= 180 J mol^-1 k^-1

Explanation / Answer

#4a

The given chemical reaction is

Mg2SiO4 (s) + 4 H2CO3 (aq) ---------------> 2 Mg^2+ (aq) + 4 HCO3^- (aq) + H4SiO4 (aq)

Let's find Delta Grxn here

Delta G^o rxn = [Delta G^o of products] - [Delta G^o of reactants]

=[ 2 * delta G^o of Mg2+ + 4* delta G^o of HCO3^- + delta G^o of H4SiO4] - [ Delta G^o of Mg2SiO4 + 4* Delta G^o of H2CO3]

= [ 2 * -455.4 + (- 4* 586.8) + (-1307.9) ] - [ (-2056.7) + (4* -623.14) ]

= [ -4565.9 ] - [ -4549.26 ]

= -16.64 kJ

( **Note : I have used negative values of delta G as delta G is often negative for most of the substances. Please check the sign of the values given to you)

Delta G^o rxn = -16.64 kJ

Delta G^o is related to Keq by following equation

Delta G^o = -R*T*ln K

16.64 kJ * 100 J / 1 kJ = 16640 J

- 16640 J = - *8.314 J/mol K * 298 K * ln K

ln K = -16640/ - 8.314*298

ln K = 6.716

K = e^ 6.716

K = 825.7

Keq for the weathering reaction at 25 C is 826

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#4b

The equilibrium reaction can be written as

Mg2SiO4 (s) + 4 H2CO3 (aq) <--------------> 2 Mg^2+ (aq) + 4 HCO3^- (aq) + H4SiO4 (aq)

When Mg2+ ions are added to the solution, we are increasing the amount of product .which means we have more Mg2+ ions in the solution. To overcome this change, equilibrium would shift towards left in order to form less Mg2+ and the reverse reaction is favored till equilibrium is reestablished.

This suggests that solubility of Mg2SiO4 would be lowered

____________________________________________________________________________________________________

#4c

Let's calculate delta H rxn for the above reaction

Delta rxn = Delta H Products - Delta H reactants

Delta H rxn = ( 2 * -467.0 + 4 * -689.93 + (-1457.3)) + ( 4 * - 699.09 + (-2175.7))

Delta H rxn = - 5151.02 - ( -4972.06)

Delta Hrxn = -178.96 kJ

( **Note : PLease check sign of delta H given in the appendix. It should be negative)

The negative value of delta H suggests that heat is evolved in this reaction which means the reaction is exothermic.

Since exothermic reactions themselves evolve heat,they are favored at lower temperatures.

If we carry out above reaction at higher temperature, reverse reaction will be favored, because forward reaction evolves heat and is therefore not favored as temperature is already high

The equilibrium would get shifted towards left

Solubility of Mg2SiO4 would decrease

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#4d

The equation we will be using to find Keq at 40 C is given below

ln ( K2/K1) = H / R ( 1/T1 - 1/T2) .............. Vant Hoff equation

here, K1 = 826

K2 = ?

T1 = 25 C + 273 = 298 K

T2 = 40 + 273 = 313 K

H = - 178.96 kJ/mol = -178960 J/mol

R = 8.314 J/mol K

Substituting above values in Vant Hoff equation, we get

ln ( K2/ 826) = (-178960 J/mol) / 8.314 J/ mol K * ( 1/ 298 K - 1/313 K )

ln ( K2/826) = -21525 * ( 0.0001608165)

ln ( K2/826) = -3.46158

K2/826 = 0.03138

K2 = 25.9

We can see from above value of K2 that as we increase the temperature , equilibrium constant decreases which indicates that less products are being formed.

Therefore solubility of Mg2SiO4 decreases as we increase the temperature

This statement is consistent with what we said in part c

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