Given reaction is:
H2(g) ---> 2H(g) .... H = 436.4 kJ/mol......(1)
Br2(g) ---> 2Br(g) ...H = 192.5 kJ/mol......(2)
H2(g) + Br2(g) ---> 2HBr (g) .... H = -72.4 kJ/mol... (3)
(3) - (1) - (2)
H2(g) + Br2(g) - H2(g) - Br2(g) ---> 2HBr (g) - 2H(g) - 2Br(g) .....H = -72.4 - 192.5 - 436.4
2H(g) + 2Br(g) ---> 2HBr(g) ... H = -701.3 kJ/mol
H(g) + Br(g) ---> HBr(g) .... H = -350.65 kJ
You are provs the Rollow g dist AH-436.4 kJ/mol AH-192.5 kJ/nol AH-72.4 kJ/mol Hod) 2 H) RMEtC (8) H) + Br() 2 HBrg)...
You are provided the following data: H2(g) → 2 H(g) Br(g) → 2 Br(g) H2(g) + Brz(g) → 2 HBr(g) AH = 436.4 kJ/mol AH = 192.5 kJ/mol AH = - 72.4 kJ/mol Calculate AH for the reaction H(g) + Br(g) → HBr(g) Show your work | What is ΔΗ?
8. A typical candy bar has 35.0 g of glucose. The human body generates energy from glucose by the thermochemical equation provided below CaH2O(s) +6 0;(g) 6 CO:(g ) + 6 H2O(0) AH 670 kcal How much energy is obtained from the digestion of a typical candy bar? Show your work Final Answer 9. You are provided the following data: H2(g) 2 H(g) Br(g) 2 Br(g) H2(g) + Br(g)2 HBr(g) AH 436.4 kJ/mol AH 192.5 kJ/mol AH-72.4 kJ/mol Calculate AH...
8. A typical candy bar has 35.0 g of glucose. The human body generates energy from glucose by the thermochemical equation provided below CH,204(s) + 6 02(g) → 6 CO2(g) + 6 H20(1) AH = -670 kcal How much energy is obtained from the digestion of a typical candy bar? Show your work Final Answer You are provided the following data: H2(g) → 2 Hg) Br(g) → 2 Br(g) H2(g) + Bra(g) → 2 HBr(g) AH = 436.4 kJ/mol AH...
Please write clearly and use sig figs. thanks in advance 8. A typical candy bar has 35.0 g of glucose. The human body generates energy from glucose by the thermochemical equation provided below C6H1206(S) + 6 O2(g) → 6 CO2(g) + 6 H2O(1) AH=-670 kcal How much energy is obtained from the digestion of a typical candy bar? Show your work Final Answer You are provided the following data: H2(g) → 2 H(g) Br(g) → 2 Br(g) H2(g) + Br2(g)...
Enter your answer in the provided box. You are given the following data: H2(g) → 2H(g) Br2(g) → 2Br(g) AH = 436.4 kJ/mol AH = 192.5 kJ/mol H2(g) + Br2(g) → 2HBr(g) AH = -72.4 kJ/mol Calculate Ahº for the reaction H(g) + Br(g) — HBr(g) kJ/mol
6. Given the following data: Brig) 2 Brig) ΔΗ'-192.5 kJ/mol H2(g) + Br2(g)-) 2HBr(g) ΔΜ"--72.4 k/mol Calculate ΔΗ' for the reaction below H(g) + Br(g) - HBr(g)
AIBr (s)Al(g)+3 Br(g) AH° = ? Chemical reaction AH in kJ/mol Type of reaction Al (g) Al (s) 329.10 Heat of sublimation Al (g) + e lonization energy (first) Al (g) 577.6 Al (g) Al (g) + e- 1816.6 lonization energy (Second) AlAl lonization energy (Third) 2744.7 (g) e Br2 (g)2 Br (g) Bond energy 193 Br (g)eBr (g) -325 Electron Affinity Al (s)+3/2 Br (g)AIBr (s) 527.2 Heat of formation
AH-116.4 kJ/mol 4. Consider the following reaction: NH.NO, (s) N20 (8) + 2 H20 (g) Calculate AGⓇ for the reaction. Substance S” (J/mol K) NH.NO, (3) 151 N:0 (8) 220 H:0 (9) 183 5. The first step in the commercial production of titanium metal is the reaction of Tio: with chlorine an graphite: TiO2 (s) + 2 Cl: (g) + 2C (5) Tici: (1) + 2 CO(g) Calculate AGº for the reaction. Substance AG? (kJ/mol Substance AG® (kJ/mol) C(s) TiCl...
A chemist measures the enthalpy change AH during the following reaction: 3 CH,Br(g) + NH3(9) (CH3),N(1) + 3 HBr(9) AH=73. kJ Use this information to complete the table below. Round each of your answers to the nearest kJ/mol, reaction ΔΗ Ok CH,Br(g) + 3NH,(8) ► ? (ch,),N(!) + 2 H Br() 2(CH3),N(1) + 6HBr(g) — 6CH,Br(g) + 2NH, () x Ś ? A ПkJ (CHR),N(1) + 3H Br(g) — 3CH,Br(g) + NH3(8) OkJ
2. The reaction N2 (g)+3 F2 (g)2 NF3 (g), has AH° is -248.6 kJ mol and ArS° is -278.7J K mol. Calculate the temperature at which the reaction will be at equilibrium. Assume A,H° and A,S are independent of temperature.