When 15.0 mL of a 8.52 E .4 M lead acetate solution is with 18.0 mL of a solutio

Question

When 15.0 mL of a 8.52 E .4 M lead acetate solution is with 18.0 mL of a solution does a participate form? For these conditions the Reaction Quotient , Q is equal to For the reaction The reaction is (reactant , product) favored order standard conditions at 274 k. The entropy change for the reaction of 1.55 moles of H2O(I) at this temperature would be The volume of water needed to dissolve 0.0605 rams Assume no volume change upon addition of the solid. For magnesium Mg, the heat of at its normal melting point of 649 degree C is 9.0 kJ/mol. The entropy change when 2.20 moles of liquid Mg freezers at 649 degree C, 1 atm is j/K

Explanation / Answer

1.

Yes, PbCO3 being highly insoluble, precipitate will form.

Pb2+ + (CO3)^2- --> PbCO3

Reaction quotient Q=1/Ksp, Ksp is solubility product of PbCO3

2.

DG > 0, so reaction is reactant favored.

DS=(DH-DG)/T for 2 moles of H2O

For 1.55 mole of H2O,

DS=1.55/2*(DH-DG)/T

=1.55/2*(571.6-482.1)/274=

0.25kJ/K

3.

Let x liter of water is taken

Ksp of Ba(CrO4) =2*10^(-10)

Molecular weight = 253.37

concentration of dissolved Ba2+ or CrO4^2- =.0608/253.37/x

Ksp=( .0609/253.37/x)^2=2*10^(-10)

X=17liter

4.

T=649+273=922K

DHf=9kj/mol=9000j/mole

Entropy change of 2.2 moles=

2.2*DHf/T=2.2*9000/922=21.5J/K

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.