Estimate the enthalpy of formation of HF from the following bond
energies: (twenty-five points)
H2 (g) + F2 (g) → 2 HF (g)
Bond: H-H =432KJ/mol
F-F=159 KJ/mol
H-F= -565 KJ/mol
Estimate the enthalpy of formation of HF from the following bond energies: (twenty-five points) H2 (g)...
Using the table of bond energies below, calculate the following enthalpy values associated with this reaction: 2 HF (g) H2 (g)F2 (g) Bond type Bond Energy (kJ/mol) 159 F-F H-H 432 H-F 565 Enter your answers to zero decimal places. Ignore significant figure rules. Include negative signs if necessary What is the sum of the reactant bond enthalpies? kJ What is the sum of the product bond enthalpies? kJ What is the overall enthalpy of reaction? kJ
The reaction of H2 with F2 produces HF with deltaH=-269kj/mol of HF. If the H-H and the H-F bond energies are 432 and 565 kj/mol, respectively, what is the F-F bondenergy?H2(g)+F2(g) ----> 2 HF(g)_________kj/mol
Using the standard enthalpy of formation data, show how the standard enthalpy of formation of HF(g) can be used to determine the bond energy. (Enter unrounded values.) (a) bond energy calculated from standard enthalpy of formation kJ/mol (b) average bond enthalpy from the bond enthalpy table k]/mol Average Bond Enthalpies bondAH bond (kJ/mol) bond AHbond (kJ/mol) bond AH bond (kJ/mol) bond AH bond (kJ/mol) O-H 0-0 467 146 495 185 203 156 364 522 335 544 413 347 614 839...
The enthalpy change for the following reaction is -137 kJ. Using bond energies, estimate the C-C bond energy in C2H6(g) СЭН4(@) + H2(@) — С2Н6(@) kJ/mol Submit Answer Retry Entire Group 7 more group attempts remaining The enthalpy change for the following reaction is -903 kJ. Using bond energies, estimate the N-H bond energy in NH3(g) 4NH3(g)502(g) ANO(g) +6H20(g) kJ/mol Submit Answer Retry Entire Group 7 more group attempts remaining The enthalpy change for the following reaction is -137 kJ....
The enthalpy change for the following reaction is 95.4 kJ. Using bond energies, estimate the N-H bond energy in N2H4(g). N2(g) + 2H2(g) N2H4(g) kJ/mol The enthalpy change for the following reaction is -92.2 kJ. Using bond energies, estimate the H-H bond energy in H2(g). 2NH3(g) N2(g) + 3H2(g) kJ/mol D Single Bonds Multiple Bonds C N O F Si P S a Br 1 H 436 413 391 463 565 318 322 347 C 413 346 305 358 485...
For the reaction: H2(g)+C2H4(g)-->C2H6(g) Bond & Bond Enthalpy H-H 436.4 kJ/mole C-H 414 kJ/mol C-C 347 kJ/mol C=C 620 kJ/mol Substance & delta Hf H2 0 C2H4 52.3 C2H6 -84.7 (a) estimate the enthalpy of reaction, using the bond enthalpy values from the table in kJ/mol (b) Calculate the enthalpy of reaction, using standard enthalpies of formation
Use bond energies to estimate the enthalpy of formation of HBr(g). BE(H–H) = 436 kJ/mol BE(Br–Br) = 192 kJ/mol BE(H–Br) = 366 kJ/mol A –52 kJ/mol B +262 kJ/mol C +104 kJ/mol D +52 kJ/mol E –104 kJ/mol
Use the given bond enthalpy data to estimate the Ho (kJ) for the following reaction. (C - H = 414 kJ, H - F = 565 kJ, C - Cl = 331 kJ, C - F = 439 kJ, F - F = 157.8 KJ). CH3Cl(g) + F2(g) =>CH2FCl(g) + HF(g)
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)
Part A. Find the average enthalpy of the PF bond in PF5(g) from the following standard molar enthalpies of formation (in kJ/mol): PF5(g) (–1595), P(g) (315), F(g) (79). Part B. Using the bond energies from the table below estimate AH for the following reaction C2H2(g) + H2(g) → C2H4(g) Bond Bond energy [kJ/mol] [C-H 413 H-H 436 |C-C 348 C=C 614 csc 839