Consider the following reaction: Part D 2 H2(g) + 2 NO(g) → N2(g) + 2 H20(g) The standard enthalpy change and standard entropy change for the reaction are: and the following table of standard enthalpies of formation and standard absolute entropies: AHP = -666.2 kJ mol-1 and A,S° = -114.0 J K-1 mol-1, respectively. compound AH (kJ mol-1) s° (J K1 mol-1) Compute the standard Gibbs energy change for the reaction, A,G°, at 32.0 °C. H2(g) 130.8 O 301.4 kJ mol-1 O 2981.8 kJ mol-1 NO(g) 91.3 210.8 O 4314.2 kJ mol-1 N2(g) 191.6 O -701.0 kJ mol-1 O -631.4 kJ mol-1 H,O(g) -241.8 188.8

Use the standard reaction enthalpies given below to determine AH°rxn for the following reaction: Given: P4(g) + 10 Cl2(g) PC15(s) -> ->> 4 PC15(s) PC13(g) + Cl2(g) P4(g) + 6 Cl2(g) → 4 PC13(g) AH°rxn = ? AH°rxn = +144 kJ AH°rxn-1 = -1298 KJ

4. Calculate the standard heat of formation of calcium carbide (CaC) from its elements in kJ/mol using the following thermochemical equations. (hint come up with the overall equation first) Ca(s) + 2H₂O ⇒ Ca(OH)2(s) + H₂(g) 2C(s) + O2(g) → 2CO CaO + H₂O⇒ Ca(OH)2() 2H2(g) + O2(g) → 2H₂O) CaO) + 3C,,→ CaCs) + CO ΔΗ°= - 414.79 kJ ΔΗ°= - 221.00 kJ ΔΗ°= - 65.19 kJ ΔΗ°= - 571.80 kJ AH°= + 462.30 kJ 1

7) Calculate the change in internal energy (AE) for a system that is giving off 25.0 kJ of heat and is changing from 12.00 L to 6.00 L in volume at 1.50 atm pressure. (Remember that 101.3 J = 1 L'atm) A) +25.9 kJ B) -16.0 kJ C) -25.9 kJ D) -24.1 kJ E) 937 kJ Answer: D Diff: 2 Page Ref: 6.4