1. During chemical weathering, forsterite is dissolved by the carbonic acid in r

Question

1. 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.

1a. Calculate the Keq for this weathering reaction at 25 °C.

1b. If the reaction is at equilibrium, using Le Chatelier’s principle, predict what would

happen if Mg2+ ions were added to the solution.

1c. Using Le Chatelier’s principle, predict what would happen to the equilibrium constant if

the reaction occurred at a higher temperature.

1d. 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

Mg2SiO4 forsterite + 4H2CO3 (aq) 2Mg2+ + 4HCO3- + H4SiO4 (aq)

1a. Calculate the Keq for this weathering reaction at 25 °C.

we need dG

dG = Gprod - Greact

dG = 455.4 + 586.8 + 1307.9 - (623.14 + 2056.7 ) = -329.74 kJ/mol

dG= -RT*ln(K)

K = exp(-dG/(RT))

K = exp(329740/(8.314*298)) = 6.3130*10^57

1b. If the reaction is at equilibrium, using Le Chatelier’s principle, predict what would happen if Mg2+ ions were added to the solution.

If we add Mg+2, which are PRODUCT, then we favour the shift wotard the LEFT, i.e. reactants, that is, MG2SiO4 (forseite production

1c. Using Le Chatelier’s principle, predict what would happen to the equilibrium constant if the reaction occurred at a higher temperature.

We must get the enthalyp of reaction,

HRxn = Hprod - Hreact

HRxn = 467.0 +689.93 +1457.3 - (2175.7 + 699.09 ) = -260.56 kJ/mol

since it is negative ,this is EXOTHERMIC

meaning that addition of Heat will not favour production!

1d. 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).

log(K1/K2)= -H/R*(1/T2 - 1/T1)

log((6.3130*10^57) / K2) = -(-260.56 *1000)/8.314*(1/(40+273) + 1/(25+273))

(6.3130*10^57) / K2)= 10^205.294

K2 = (6.3130*10^57) / 10^205.294 = 3.208*10^-148

since this is EXOTHERMIC, an increase in T is not favoured

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