Please explain
2 ClF (g) + O2 (g) ? Cl2O (g) + F2O (g) ?H = 167.4 kJ
2 ClF3 (g) + 2 O2 (g) ? Cl2O (g) + 3 F2O (g) ?H = 341.4 kJ
2 F2 (g) + O2 (g) ? 2 F2O (g) ?H = -43.4 kJ
Calculate the ?H for the reaction
ClF (g) + F2 (g) ? ClF3 (g)
Homework Answers
At 25°C, the following heats of reaction are known: 2CIF(g) + O2(g) → Cl2O(g) + F2O(g)...
At 25°C, the following heats of reaction are known: 2CIF(g) + O2(g) → Cl2O(g) + F2O(g) AHºrn = 167.4 kJ/mol 2C1F3(g) + 202(g) + Cl2O(g) + 3F2O(g) AHºrn = 341.4 kJ/mol 2F2(g) + O2(g) → 2F20(g) AHºrxn=-43.4 kJ/mol At the same temperature, use the above data to calculate the heat released (kJ) when 3.40 moles of CIF(g) reacts with excess F2. CIF(g) + F219) -> CIF3(g) Write answer to three significant figures. NO SIGN in ANSWER. Numeric Response
AH° = ? 3.(5pts) Consider the thermochemical reaction: CIF(g) + F (g) ----> CIF3(g) Use the...
AH° = ? 3.(5pts) Consider the thermochemical reaction: CIF(g) + F (g) ----> CIF3(g) Use the following data to determine the value of the missing quantity in the above equation. 2 CIF(g) + O2(g) -----> CI2O(g) + F2O(g) 2 CIF,(g) + 2 O2(g) ----> CI-O(g) + 3 F2O(g) 2 F2(g) + O2(g) -- -> 2 F.0(g) AH AH AH = 167.4 kJ/mol = 341.4 kJ/mol = -43.4 kJ/mol
9. Given the following data: 2C1F(g) + O2(g) → Cl2O(g) + F20(g) 2C1F3(g) + 2O2(g) → Cl2O(g) + 3F20 2F2(g) + O2(g)...
9. Given the following data: 2C1F(g) + O2(g) → Cl2O(g) + F20(g) 2C1F3(g) + 2O2(g) → Cl2O(g) + 3F20 2F2(g) + O2(g) → 2F20(g) AH = 167.4 kJ AH = 341.4 kJ AH = -43.4 kJ Calculate AH for the reaction CIF(g) + F2(g) → CIF3(g) 10. Calculate AHº for the following reaction using the AHⓇ information below. 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) AH NH3(g) is -80 kJ/mol AH° NO(g) is 90 kJ/mol AH®, H2O(g) is -242 kJ/mol
pter 4-6 0 Saved Help Save At 25°C, the following heats of reaction are known: 2C1F(g)...
pter 4-6 0 Saved Help Save At 25°C, the following heats of reaction are known: 2C1F(g) + O2(g) → Cl2O(g) +F20(g) 2C1F3(g) + 2O2(g) + Cl2O(g) + 3F2O(g) 2F2(g) + O2(g) → 2F20(g) AHO/= 167.4 kJ/mol AHºrn = 341.4 kJ/mol AHºrn=-43.4 kJ/mol At the same temperature, use the above data to calculate the heat released (kJ) when 3.40 moles of CIF(g) reacts with excess F2. CIF(g) + F2(9) --> CIF3(9) Write answer to three significant figures. NO SIGN in ANSWER....
At 25°C, the following heats of reaction are known: 2CIF(g) + O2(g) → Cl20(g) + F2。dH'm...
At 25°C, the following heats of reaction are known: 2CIF(g) + O2(g) → Cl20(g) + F2。dH'm = 167.4 kJ/mol 2ClF3(g) + 202(g) → Cl20(g) + 3F20(g) ΔH'm-34 1.4 kJ/mol 2F2(g) + O2(g) → 2F20(g) ΔHoxn--43.4 kJ/mol At the same temperature, use Hess's law to calculate ΔHorn for the reaction: CIF(g) + F2(g) → ClF3(g) 0 a-21 7.5 kJ/mol O b 465.4 kJ/mol O c. -108.7 kJ/mol d 217.5 kJ/mol O e-130.2 kJ/mol
At 25 degree C, the following heats of reaction are known: 2CIF (g) + O_2 (g)...
At 25 degree C, the following heats of reaction are known: 2CIF (g) + O_2 (g) rightarrow Cl_2O (g)+ F_2O (g) Delta H_rxn^degree = 167.4 kJ/mol 2ClF_3 (g) +2O_2 (g) rightarrow Cl_2O (g) + 3F_2O (g) Delta H_rxn^degree = 341.4 kJ/mol 2F_2 (g) + O_2 (g) rightarrow 2F_2O (g) Delta H_rxn^degree = -43.4 kJ/mol At the same temperature, use Hess' law to calculate Delta H_rxn^degree for the following reaction: ClF (g) + F_2 (g) rightarrow ClF_3 (g)
At 25°C, the following heats of reaction are known: AH (kJ/mol 167.4 2CIF + 02 →Cl20...
At 25°C, the following heats of reaction are known: AH (kJ/mol 167.4 2CIF + 02 →Cl20 + F20 2ClF3 + 202 →Cl20+3F20 341.4 2F2 + 02 → 2F20 At the same temperature, calculate ΔH for the reaction: ClF + F2 → CIF3 -43.4 A. -217.5 kJ/mol B.-130.2 kJ/mol C. +217.5 kJ/mol ○ D.-108.7 kJ/mol E. none of these QUESTION 4 Consider the reaction: When a 12.9-g sample of ethyl alcohol(molar mass 46.07 g/mol) is burned, how much energy is released...
3.(5pts)) Consider the thermochemical reaction: 3 CIF(g) + 3F.(g) ---->3 CIF,(g) AHºrn = ? Use the...
3.(5pts)) Consider the thermochemical reaction: 3 CIF(g) + 3F.(g) ---->3 CIF,(g) AHºrn = ? Use the following data to determine the value of the missing quantity in the above equation. 6 CIF(g) + 3 O2(g) ----->3 ClO(g) + 3 F 0(g) 2 CIF3(g) + 2 02(8) -----> CLO(g) + 3 F.0() 2 F2(g) + O2(g) -------> 2 F.O(g) AH AH° AH - 502.2 kJ/mol 341.4 kJ/mol = -43.4 kJ/mol
Part A. Given the bond dissociation energies (in kJ/mol) for the following diatomic molecules Cl2 (243),...
Part A. Given the bond dissociation energies (in kJ/mol) for the following diatomic molecules Cl2 (243), F2 (158), H2 (436), O2 (498), N2 (945) choose the one(s) that could be broken by using blue light (λ=465 nm). Part B. Given the bond energies (in kJ/mol) of the following bonds: F–F (155), F–Cl (193), and Cl–Cl (243), estimate the molar enthalpy of formation of ClF(g), that is find ∆H for the following reaction ½Cl2(g) + ½F2(g) → ClF(g)
Given the bond energies (in kJ/mol) of the following bonds: F–F (155), F–Cl (193), and Cl–Cl (243), estimate the molar e...
Given the bond energies (in kJ/mol) of the following bonds: F–F (155), F–Cl (193), and Cl–Cl (243), estimate the molar enthalpy of formation of ClF(g), that is find ∆H for the following reaction ½Cl2(g) + ½F2(g) → ClF(g) A. 209 kJ/mol B. -50. kJ/mol C. –8 kJ/mol D. –209 kJ/mol E. 8 kJ/mol
At 25°C, the following heats of reaction are known: 2CIF(g) + O2(g) → Cl2O(g) + F2O(g) AHºrn = 167.4 kJ/mol 2C1F3(g) + 202(g) + Cl2O(g) + 3F2O(g) AHºrn = 341.4 kJ/mol 2F2(g) + O2(g) → 2F20(g) AHºrxn=-43.4 kJ/mol At the same temperature, use the above data to calculate the heat released (kJ) when 3.40 moles of CIF(g) reacts with excess F2. CIF(g) + F219) -> CIF3(g) Write answer to three significant figures. NO SIGN in ANSWER. Numeric Response
AH° = ? 3.(5pts) Consider the thermochemical reaction: CIF(g) + F (g) ----> CIF3(g) Use the following data to determine the value of the missing quantity in the above equation. 2 CIF(g) + O2(g) -----> CI2O(g) + F2O(g) 2 CIF,(g) + 2 O2(g) ----> CI-O(g) + 3 F2O(g) 2 F2(g) + O2(g) -- -> 2 F.0(g) AH AH AH = 167.4 kJ/mol = 341.4 kJ/mol = -43.4 kJ/mol
9. Given the following data: 2C1F(g) + O2(g) → Cl2O(g) + F20(g) 2C1F3(g) + 2O2(g) → Cl2O(g) + 3F20 2F2(g) + O2(g) → 2F20(g) AH = 167.4 kJ AH = 341.4 kJ AH = -43.4 kJ Calculate AH for the reaction CIF(g) + F2(g) → CIF3(g) 10. Calculate AHº for the following reaction using the AHⓇ information below. 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) AH NH3(g) is -80 kJ/mol AH° NO(g) is 90 kJ/mol AH®, H2O(g) is -242 kJ/mol
pter 4-6 0 Saved Help Save At 25°C, the following heats of reaction are known: 2C1F(g) + O2(g) → Cl2O(g) +F20(g) 2C1F3(g) + 2O2(g) + Cl2O(g) + 3F2O(g) 2F2(g) + O2(g) → 2F20(g) AHO/= 167.4 kJ/mol AHºrn = 341.4 kJ/mol AHºrn=-43.4 kJ/mol At the same temperature, use the above data to calculate the heat released (kJ) when 3.40 moles of CIF(g) reacts with excess F2. CIF(g) + F2(9) --> CIF3(9) Write answer to three significant figures. NO SIGN in ANSWER....
At 25°C, the following heats of reaction are known: 2CIF(g) + O2(g) → Cl20(g) + F2。dH'm = 167.4 kJ/mol 2ClF3(g) + 202(g) → Cl20(g) + 3F20(g) ΔH'm-34 1.4 kJ/mol 2F2(g) + O2(g) → 2F20(g) ΔHoxn--43.4 kJ/mol At the same temperature, use Hess's law to calculate ΔHorn for the reaction: CIF(g) + F2(g) → ClF3(g) 0 a-21 7.5 kJ/mol O b 465.4 kJ/mol O c. -108.7 kJ/mol d 217.5 kJ/mol O e-130.2 kJ/mol
At 25 degree C, the following heats of reaction are known: 2CIF (g) + O_2 (g) rightarrow Cl_2O (g)+ F_2O (g) Delta H_rxn^degree = 167.4 kJ/mol 2ClF_3 (g) +2O_2 (g) rightarrow Cl_2O (g) + 3F_2O (g) Delta H_rxn^degree = 341.4 kJ/mol 2F_2 (g) + O_2 (g) rightarrow 2F_2O (g) Delta H_rxn^degree = -43.4 kJ/mol At the same temperature, use Hess' law to calculate Delta H_rxn^degree for the following reaction: ClF (g) + F_2 (g) rightarrow ClF_3 (g)
At 25°C, the following heats of reaction are known: AH (kJ/mol 167.4 2CIF + 02 →Cl20 + F20 2ClF3 + 202 →Cl20+3F20 341.4 2F2 + 02 → 2F20 At the same temperature, calculate ΔH for the reaction: ClF + F2 → CIF3 -43.4 A. -217.5 kJ/mol B.-130.2 kJ/mol C. +217.5 kJ/mol ○ D.-108.7 kJ/mol E. none of these QUESTION 4 Consider the reaction: When a 12.9-g sample of ethyl alcohol(molar mass 46.07 g/mol) is burned, how much energy is released...
3.(5pts)) Consider the thermochemical reaction: 3 CIF(g) + 3F.(g) ---->3 CIF,(g) AHºrn = ? Use the following data to determine the value of the missing quantity in the above equation. 6 CIF(g) + 3 O2(g) ----->3 ClO(g) + 3 F 0(g) 2 CIF3(g) + 2 02(8) -----> CLO(g) + 3 F.0() 2 F2(g) + O2(g) -------> 2 F.O(g) AH AH° AH - 502.2 kJ/mol 341.4 kJ/mol = -43.4 kJ/mol
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