Answer to Question #98952 in General Chemistry for Brittany Wallace

Question

1. The enthalpy change for the following reaction is 81.1 kJ. Using bond energies, estimate the H-Cl bond energy in HCl(g).

2HCl(g) + Br2(g)2HBr(g) + Cl2(g)

___ kJ/mol

2. The enthalpy change for the following reaction is -137 kJ. Using bond energies, estimate the C-C bond energy in C2H6(g).

C2H4(g) + H2(g)C2H6(g)

___ kJ/mol

Answer 1

1. $2HCl(g) + Br_2(g) \rightarrow 2HBr(g) + Cl_2(g)$ $\Delta H_{rxn} = 81.1 kJ$

$$\Delta H _{rxn} = \sum \Delta H(bonds broken) - \sum \Delta H (bonds formed)$$

Bonds energies:

Br-Br 193 kJ/mol

H-Br 366 kJ/mol

Cl-Cl 242 kJ/mol

$$81.2 = [2\times(H-Cl)+1\times (Br-Br)] - [2\times(H-Br)+1\times(Cl-Cl)]$$

$$81.2 = [2\times(H-Cl)+193]-[2\times366+242]$$

Bond energy of H-Cl is 431.1 kJ/mol

2.

$C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g)$ $\Delta H_{rxn} = -137 kJ$

$$\Delta H_{rxn} = \sum \Delta H (bonds broken) - \sum \Delta H(bonds formed)$$

Bonds energies:

C=C 614 kJ/mol

C-H 413 kJ/mol

H-H 436 kJ/mol

$-137 = [1\times (C=C)+ 4\times(C-H)+1\times(H-H)]-[1\times(C-C)+6\times(C-H)]$

$-137 = [614+4\times413+436]-[(C-C)+6\times413]$

Bond enrgy of C-C is: 361 kJ/mol