(6 pts.) The standard enthalpy of sublimation of phenanthrene is Hsub-841 kJ/mol

Question

(6 pts.) The standard enthalpy of sublimation of phenanthrene is Hsub-841 kJ/mol at 298 K. What is the standard internal energy change of sublimation Usub at this temperature? 4, AHub 84.1 kJ/mol 5. (10 pts.) Find the enthalpy of reaction for the reaction CS2(I) + 3 O2 (g) CO2(g) + 2 SO2(g) given the following reactions and enthalpies C(s) + O2(g) CO2(g) S(s) +02(g) SO2(g) C(s) + 2 S(s) CS2(1) H =-393.5 kJ H =-296.8 kJ AH-87.9 kJ (8 pts.) Starting with U(V,T), and using the method discussed in class, derive the expression 6. (ov),-(au),(av)

Explanation / Answer

Q4

dH = dU + D(PV)

at constant P, assume

dH = DU + P*dV

dU = dH - P*dV

P = 101325 Pa, dV = Vfinal - Vinitial = Vgas - Vliquid ---> Vgas approx

1 mol will occupy --> 22.4 L at 0°C ---> V(298) = 22.4*298/273 = 24.451 L

now...

dU = dH - P*dV= (84.1*10^3) - (101325)(24.451*10^-3) =

dU = 81622.502 kJ

Q5.

HRxn can be obtained via hess law:

invert rxn 3 since we need CS2 in the left

CS2 = C + 2S H = -87.9

multiply SO2 by 2

2S + 2O2 = 2SO2 H = -296.8*2 = -593.6

now, add all

C + O2 + CS2 + 2S + 2O2 = 2SO2+C + 2S +CO2 H = -593.6 -87.9 - 393.5 = -1075

cancel common terms

CS2 + 3O2 = 2SO2+ CO2 H = -1075

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