Homework Answers
Given:
Hof(P(g)) = 315.0 KJ/mol
Hof(F(g)) = 79.0 KJ/mol
Hof(PF5(g)) = -1595.0 KJ/mol
Balanced chemical equation is:
P(g) + 5 F(g) ---> PF5(g)
ΔHo rxn = 1*Hof(PF5(g)) - 1*Hof( P(g)) - 5*Hof(F(g))
ΔHo rxn = 1*(-1595.0) - 1*(315.0) - 5*(79.0)
ΔHo rxn = -2305 KJ
There are no bonds in reactant and there are 5 P-F bonds in product
ΔHo rxn can also be written as:
ΔHo rxn = BE(reactant) - BE(products)
-2305 = 0 - 5*BE(P-F)
5*BE(P-F) = 2305
BE(P-F) = 461 KJ/mol
Answer: 461 KJ/mol
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Question 6 1 pts Find the average enthalpy of the PF bond in PFs(g) from the following standard molar enthalpies of for...
Part A. Find the average enthalpy of the PF bond in PF5(g) from the following standard...
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
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Using the bond enthalpies in the Average Bond Enthalpies table, determine the approximate enthalpy (in kJ)...
Using the bond enthalpies in
the Average Bond Enthalpies table, determine the approximate
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ITITIT Average Bond Enthalpies AH bond (kJ/mol) bond AHond (kJ/mol) bond AH bond (kJ/mol) bond bond AH bond (kJ/mol) С-Н...
Using the bond enthalpies in the Average Bond Enthalpies table, determine the approximate enthalpy (in kJ)...
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If you look in the tables of bond enthalpies in a textbook (or most other places)...
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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
Using the bond enthalpies in
the Average Bond Enthalpies table, determine the approximate
enthalpy (in kJ) for each of the following reactions. (Assume the
average bond enthalpy of the Cl–F bond is 254 kJ/mol.) (a) Cl2(g) +
3 F2(g) → 2 ClF3(g) (b) H2C=CH2(g) + H2(g) → H3CCH3(g) (c) 2
CH3(C=O)H(g) + 5 O2(g) → 4 CO2(g) + 4 H2O(g)
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Use the molar bond enthalpy data in the table to estimate the Average molar bond enthalpies. (Hd value of AHn for the equation kJ mol Bond Воnd kJ mol CH)+HBrig)C,H,Brig) O-H 464 CEN 890 O-O The bonding in the molecules is shown. 142 N-H 390 C-O 351 N-N 159 H Br O-O 502 N-N 418 H-Br NEN C-O 730 945 C-C 347 F-F 155 Enter numeric value С-С 615 CI-CI 243 AH = C C Br-Br I kJ 811 192...
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If you look in the tables of bond enthalpies in a textbook (or most other places) you will not find entries for bonds for any noble gas elements. Given the equation below for the formation of XeF. Xe (g) + 2F2(g) + XeF4(g) AH= -281 kJ/mol and the fact that the average F-F bond enthalpy is 156 kJ/mol, calculate the average Xe-F bond enthalpy in XeF4 0-593 kJ/mol 593 kJ/mol • 148 kJ/mol 156 kJ/mol 109 kJ/mol
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