Enter your answer in the provided box. You are given the following data: H2(g) → 2H(g)...
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?
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)
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)...
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...
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) Caleulate At Ror the reaction (ह)।{।। - (त)।।। + ()। Show your work What is AH?
5. For the overall exothermic reaction: H2(g) + Br2(g) → 2HBr(g) the following mechanism was determined: fast Equilibrium Step 1: Brz(8) 2Br(g) Step 2: H2(g) + Br(g)_52HBr(g) + H(g) Step 3: H(g) + Br(g) k3 HBr(g) slow fast Use a plot of AH versus Reaction Pathway to illustrate the three step reaction profile.
Consider the reaction: 2HBR(g) >H2(g) + Br2() Using standard thermodynamic data at 298K, calculate the free energy change when 1.51 moles of HBr(g) react at standard conditions AG° kJ rxn AHof (kJ/mol) AG°F (kJ/mol) s° (J/mol K) Beryllium Вe(s) 0 9.5 -569.0 ВeO(s) -599.0 14.0 Be(ОН)2(s) -902.5 -815.0 51.9 AH°f (kJ/mol) AG°f (kJ/mol) s° (J/mol K) Bromine Br(g) 111.9 175.0 82.4 Br2() 152.2 0 0 Br2(g) 30.9 3.1 245.5 Br2(aq) -3.0 4.0 130.0 -121.0 -175.0 82.0 Br (aq) BrF3(g) -255.6...
Enter your answer in the provided box. Given the reaction: H2(g) + I2(g) ⇌ 2HI(g) ΔG o for the reaction is 2.60 kJ/mol at 25degree C. What is the minimum partial pressure of I2 required for the reaction to be spontaneous in the forward direction at 2degrees C if the partial pressures of H2 and HI are 3.5 and 1.75 atm, respectively?
Enter your answer in the provided box From the following data, C(graphite) + O2(0)+ CO2(g) An° . =-393.5 kJ/mol rxn Hy(@) +0,6) H200 AH =-285.8 kJ/mol rxn 2C2H6(8) + 1026) →40026) + 6H20(1) AH =-3119.6 kJ/mol rxn calculate the enthalpy change for the reaction below: 2 C(graphite) + 3H2(g) → CH()