To estimate the value of Δ𝐻∘rxn for the given equation, we need to calculate the total bond enthalpy of the bonds broken minus the total bond enthalpy of the bonds formed.
Bonds broken:
1. C-C bond in C2H4: C-C bond enthalpy = 347 kJ/mol (2 C-C bonds broken)
2. H-Br bond in HBr: H-Br bond enthalpy = 368 kJ/mol
Bonds formed:
1. C-C bond in C2H5Br: C-C bond enthalpy = 347 kJ/mol
2. C-Br bond in C2H5Br: C-Br bond enthalpy = 276 kJ/mol
Total bond enthalpies of bonds broken = (2 * 347 kJ/mol) + 368 kJ/mol = 1062 kJ/mol
Total bond enthalpies of bonds formed = 347 kJ/mol + 276 kJ/mol = 623 kJ/mol
Δ𝐻∘rxn = Total bond enthalpies of bonds broken - Total bond enthalpies of bonds formed
Δ𝐻∘rxn = 1062 kJ/mol - 623 kJ/mol = 439 kJ/mol
Therefore, the estimated value of Δ𝐻∘rxn for the given equation is 439 kJ/mol.
Use the molar bond enthalpy data in the table to estimate the value of Δ𝐻∘rxn for the equation
C2H4(g)+HBr(g)⟶C2H5Br(g)
The bonding in the molecules is shown.
Two carbon atoms are connected by a double line. Each carbon atom is connected to two hydrogen atoms through single lines. Plus sign. A hydrogen atom is connected to a bromine atom through a single line. Reaction arrow. Two carbon atoms are connected by a single line. One carbon atom is connected to three hydrogen atoms by single lines. The other carbon atom is connected to two hydrogen atoms and one bromine atom through single lines.
Average molar bond enthalpies. (𝐻bond)
Bond 𝐤𝐉⋅𝐦𝐨𝐥−1 Bond 𝐤𝐉⋅𝐦𝐨𝐥−1
O−H 464 C≡N 890
O−O 142 N−H 390
C−O 351 N−N 159
O=O 502 N=N 418
C=O 730 N≡N 945
C−C 347 F−F 155
C=C 615 Cl−Cl 243
C≡C 811 Br−Br 192
C−H 414 H−H 435
C−F 439 H−F 565
C−Cl 331 H−Cl 431
C−Br 276 H−Br 368
C−N 293 H−S 364
C=N 615 S−S 225
1 answer