As a result of mineral erosion and biological activity, phosphate ion is common

Question

As a result of mineral erosion and biological activity, phosphate ion is common in natural waters, where it often precipitates as insoluble salts, such as Ca_3(PO_4)_2. If [Ca^2+] = [PO_4^3-] = 1.0 times 10^-7 M in a given river will Ca_3(PO_4)_2: precipitate? K_sp of Ca_3(PO_4)_2 is 1.2 times 10^-29. Use data in Appendix C to calculate delta H degree, delta S degree and delta G degree at 25 degree C for each of the following reaction. In each case show that delta G degree = delta H degree - T delta S degree. 2 CH_3OH(l) + 3O_2(g) rightarrow 2 CO_2(g) + 4H_2O(l) P_4O_10(s) + 6H_2O(l) rightarrow 4H_3PO_4(aq) Using data from Appendix C, write the equilibrium-constant expression and calculate the value of the equilibrium constant and the free energy for the following reactions N_2O_4(g) = 2NO_2(g) BaCO_3(s) = BaO(s) + CO_2(g)

Explanation / Answer

Ca3(PO4)2 ---> 3Ca2+ + 2PO4 3-

Q= reaction coefficient = [Ca+2]3 [PO4-3]2

given [Ca+2] =1*10-7= [PO4-3]

Q = (3*10-7)3*(2*10-7)2= 1.08*10-33, since Q<KSp, no precipitate forms

2.

deltaH= 2* standard enthalpy of formation of CO2+4* standard enthalpy of formation of H2O- {2*standard enthalpy of formation CH3OH+3* standard enthalpy of formation of O2 (1)

standard enthalpy of formation ( Kj/mole) : CH3OH= -239, CO2(g) :-394, H2O(l):-286, O2=0

deltaH= 2*-394+4*(-286)-{2*(-239)+0}=-1454 Kj

Similar to standard enthalpy change,

Standard entropy change= 2* standard entropy   of formation of CO2+4* standard entropy of formation of H2O- {2*standard entropy of formation CH3OH+3* standard enttropy of formation of O2}

2,4, 2 and 3 are coefficients of CO2, H2O, CH3OH and O2 respectivley in the reaction.

Standard entropy data (J/mole.K) : CH3OH=126.8, O2= 205.8, CO2 = 213.6, H2O(l)=69.9

Standard entropy change= 2*213.6+4*69.9-(2*126.8+3*205.8)=-164.2 J/mole.K

deltaG= deltaH-T*deltas= -1454+298*164.2/1000 Kj=-1405.07 Kj

3.

Standard enthalpy change= 4* standard enthalpy change of H3PO4-{standard enthalpy change of P4O10+ 6* standard enthalpy change of H2O(l)}

=4*( -1277) – {6*(-286)+(-3012.5)}=-379.5 Kj

Entropy change= 4*(-221.75) –{ 6*69.9+1*228.9)=-1535.3 J/mole.K

deltaH= -379.5+2*1535.3/1000 = -376.4 Kj/mole

4. for the reaction N2O4-->2NO2, K= Equilibrium constant = [NO2]2/ [N2O4]=0.212 at 100 deg.c

deltaG=-RT ln K= -8.314*373*ln(0.212)= 4810.36 Kj/mole

for the reaction , BaCO3(s)----->BaO(s)+CO2(g), K= [CO2]

deltaG= 1*(-393.51)+1*(-548.1)-{1*-1132} = 190.6 Kj/mole

deltaG=-RT lnK= 190.6

lnK= -190.6*1000/(8.314*298), K= -3.88*10-34

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