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
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.
To estimate the value of Δ𝐻∘rxn for the given equation, we need to calculate the sum of the bond enthalpies for the bonds broken minus the sum of the bond enthalpies for the bonds formed.
Let's analyze the bonds broken and formed in the equation:
Bonds Broken:
1. C=C bond in C2H4 (2 x 615 kJ/mol)
2. H-Br bond in HBr (1 x 368 kJ/mol)
Bonds Formed:
1. C-Br bond in C2H5Br (1 x 276 kJ/mol)
2. C-H bond in C2H5Br (5 x 414 kJ/mol)
Now, let's calculate the bond enthalpy changes:
Bonds Broken:
(2 x 615 kJ/mol) + (1 x 368 kJ/mol) = 1598 kJ/mol
Bonds Formed:
(1 x 276 kJ/mol) + (5 x 414 kJ/mol) = 2466 kJ/mol
Finally, we can calculate the estimated value of Δ𝐻∘rxn:
Δ𝐻∘rxn = (Bonds Broken) - (Bonds Formed)
Δ𝐻∘rxn = 1598 kJ/mol - 2466 kJ/mol
Δ𝐻∘rxn ≈ -868 kJ/mol
Therefore, the estimated value of Δ𝐻∘rxn for the given equation is approximately -868 kJ/mol.
To estimate the value of Δ𝐻∘rxn for the equation C2H4(g) + HBr(g) ⟶ C2H5Br(g), we need to calculate the sum of the bond enthalpies for the reactants and subtract the sum of the bond enthalpies for the products.
Reactants:
C2H4(g) bonds:
- Two C−C bonds = 347 kJ/mol x 2 = 694 kJ/mol
- Four C−H bonds = 414 kJ/mol x 4 = 1656 kJ/mol
HBr(g) bonds:
- One H−Br bond = 368 kJ/mol
Total bond enthalpy for reactants = 694 kJ/mol + 1656 kJ/mol + 368 kJ/mol = 2718 kJ/mol
Products:
C2H5Br(g) bonds:
- One C−C bond = 347 kJ/mol
- Five C−H bonds = 414 kJ/mol x 5 = 2070 kJ/mol
- One C−Br bond = 276 kJ/mol
Total bond enthalpy for products = 347 kJ/mol + 2070 kJ/mol + 276 kJ/mol = 2693 kJ/mol
Δ𝐻∘rxn = Total bond enthalpy of reactants - Total bond enthalpy of products
Δ𝐻∘rxn = 2718 kJ/mol - 2693 kJ/mol
Δ𝐻∘rxn ≈ 25 kJ/mol
Therefore, the estimated value of Δ𝐻∘rxn for the equation C2H4(g) + HBr(g) ⟶ C2H5Br(g) is approximately 25 kJ/mol.