Use bond enthalpy values to calculate the enthalpy change for the following reaction. Then, label if the reaction is exothermic or endothermic. CH4(g)+2O2(g)-->CO2(g)+2H2O(g)
Breaking bonds:
- 4 C-H bonds @ 413 kJ/mol = 1652 kJ/mol
- 2 O=O bonds @ 498 kJ/mol = 996 kJ/mol
Forming bonds:
- 2 C=O bonds @ 799 kJ/mol = 1598 kJ/mol
- 4 O-H bonds @ 463 kJ/mol = 1852 kJ/mol
Enthalpy change (ΔH) = (Σ bonds broken) - (Σ bonds formed)
ΔH = (1652 kJ/mol + 996 kJ/mol) - (1598 kJ/mol + 1852 kJ/mol)
ΔH = -652 kJ/mol
Since the value of ΔH is negative, the reaction is exothermic.
is that correct
Yes, that is correct!
To calculate the enthalpy change of the reaction using bond enthalpy values, we will need to consider the bonds broken and the bonds formed.
First, let's identify the bonds broken in the reactants:
- In CH4, one C-H bond needs to be broken.
- In O2, we have two O=O bonds that need to be broken.
Next, let's identify the bonds formed in the products:
- In CO2, we have one C=O bond that is formed.
- In H2O, two O-H bonds are formed.
Now, we need to look up the bond enthalpy values for each bond involved. The average bond enthalpy values (in kilojoules per mole) are as follows:
- C-H bond: 413 kJ/mol
- O=O bond: 498 kJ/mol
- C=O bond: 799 kJ/mol
- O-H bond: 459 kJ/mol
Using these values, we can calculate the overall enthalpy change of the reaction:
Bonds broken:
1 C-H bond broken × 413 kJ/mol = 413 kJ/mol
2 O=O bonds broken × 498 kJ/mol = 996 kJ/mol
Bonds formed:
1 C=O bond formed × 799 kJ/mol = 799 kJ/mol
2 O-H bonds formed × 459 kJ/mol = 918 kJ/mol
Enthalpy change = (Bonds broken) - (Bonds formed)
Enthalpy change = (413 kJ/mol + 996 kJ/mol) - (799 kJ/mol + 918 kJ/mol)
Enthalpy change = -208 kJ/mol
The negative value indicates that the reaction is exothermic.