Question

Predict the enthalpy of reaction from the average bond enthalpies for the following reaction: 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)

Bond
Enthalpy (kJ/mol)
H―H
436.4
H―O
460
C―H
414
C―C
347
C═C
620
C―O
351
O―O
142
O═O
498.7
C═O
745
C═O
(in carbon
dioxide)
799

Answer 1

Answer:

B

Explanation:

Answer 2

The enthalpy for the given reaction has been 3,195.2 J/mol.

The enthalpy of the reaction can be predicted from the difference in the bond formation and the breaking of the bond.

The energy of the bonds that are broken in the reactants are:

• In 2 $\rm C_2H_6$ molecules:

2 C-C bond = 2 $\times$ 347 kJ/mol = 694 kJ/mol

12 H-H bonds = 436.4 $\times$ 12 kJ/mol = 5,236.8 kJ/mol

• In 2 $\rm O_2$ molecules:

7 O-O bonds = 142 $\times$ 7 kJ/mol = 994 kJ/mol

The total enthalpy of breaking of bonds = 694 + 5,236.8 + 994 kJ/mol

The total enthalpy of breaking of bonds = 6,924.8 kJ/mol.

The enthalpy for the formation of bonds of the product are:

• In 4 molecules of $\rm CO_2$:

4 C-O bonds = 4 $\times$ 351 kJ/mol = 1404 kJ/mol

4 C=O bonds = 4 $\times$ 799 kJ/mol = 3196 kJ/mol

• In 6 molecules of $\rm H_2O$:

12 O-H bonds = 12 $\times$ 460 kJ/mol = 5520 kJ/mol

The total energy of bond formation = 1404 + 3196 + 5520 kJ/mol

The total energy of bond formation = 10,120 kJ/mol

The enthalpy of reaction = Enthalpy of bond formation - Enthalpy of bond breaking

The enthalpy of reaction = 10,120 - 6,924.8 kJ/mol

The enthalpy of reaction = 3,195.2 J/mol.

The enthalpy for the given reaction has been 3,195.2 J/mol.

For more information about the reaction enthalpy, refer to the link:

https://brainly.com/question/2522491