Question

Table 1: Average Bond Energies (kJ/mol)

Bond	Energy	Bond	Energy
H-H	432	F-F	154
H-F	565	O-H	467
C-H	413	Cl-Cl	239
C-O	358	C=C	614
C=	1072	C=O	745
C-C	347	C=O(for $C{O_2}$(g))	799
C-Cl	339	O=O	495

Calculate the enthalpy change from bond energies for each of these reactions :
1. ${H_2}(g) + {F_2}(g) \to 2HF(g) \,\,$, $\Delta H$= __________
2. $C{H_4}(g) + 2{O_2}(g) \to C{O_2}(g) + 2{H_2}O(g) \,\,$, $\Delta H$= __________
3. $CO(g) + 2{H_2}(g) \to C{H_3}OH(l) \,\,$, $\Delta H$= __________

Answer 1

Hint: The enthalpy change may be defined as the subtraction of enthalpy of products from the enthalpy of reactants. It is the change in energy that occurred during reaction. So, the enthalpy change = Total energy of reactants - Total energy of products

Complete answer:
The bond energy is the amount of energy that is required to break one mole of that bond.
So, we have been given bond energies of various bonds. By addition of various bond energies, we can get the value of sum of energy of reactants and similarly for products. Then by subtracting, we can find the change in enthalpy.
The first reaction is-
${H_2}(g) + {F_2}(g) \to 2HF(g)$
In this, one mole of H-H and one mole of F-F bond need to be broken. And there is formation of two moles of H-F bond.
So, the bond energy for H-H bond = 432 kJ/mol
the bond energy for F-F bond = 154 kJ/mol
Total energy of reactants = 432 + 154 = 586 kJ/mol
bond energy for H-F bond = 565 kJ/mol
bond energy for two mole of H-F bond = 2$ \times $565 kJ/mol
Total energy of products = bond energy for two mole of H-F bond = 1130 kJ/mol
Change in enthalpy = 586 - 1130
Change in enthalpy = -544 kJ/mol

Similarly for the second reaction-
$C{H_4}(g) + 2{O_2}(g) \to C{O_2}(g) + 2{H_2}O(g)$
the bond energy for one C-H bond = 413 kJ/mol
the bond energy for four C-H bond = 1652 kJ/mol
the bond energy for O=O bond = 495 kJ/mol
the bond energy for two O=O bond = 990 kJ/mol
Total energy of reactants = 1652 + 990 = 2642 kJ/mol
bond energy for C=O bond = 799 kJ/mol
bond energy for two C=O bond = 1598 kJ/mol
bond energy for H-O bond = 467 kJ/mol
bond energy for four H-O bond = 1868 kJ/mol
Total energy of products = bond energy for four H-O bond = 1868 kJ/mol
Change in enthalpy = 2642 - 1868
Change in enthalpy = 774 kJ/mol

Similarly for the third reaction-
$CO(g) + 2{H_2}(g) \to C{H_3}OH(l)$
the bond energy for one C-O bond = 358 kJ/mol
the bond energy for H-H bond = 432 kJ/mol
the bond energy for two H-H bond = 864 kJ/mol
Total energy of reactants = 358 + 864 = 1222 kJ/mol
bond energy for C-H bond = 413 kJ/mol
bond energy for three C-H bond = 1239 kJ/mol
bond energy for C-O bond = 358 kJ/mol
bond energy for H-O bond = 467 kJ/mol
Total energy of products = 1239 + 358 + 467 = 2064 kJ/mol
Change in enthalpy = 1222 - 2064
Change in enthalpy = -842 kJ/mol

Note:
It must be noted that in case, if two moles of a reactant is used or two moles of product are formed then energy will be doubled. If the enthalpy change is negative, meaning energy of product is more. So, this means the reaction is exothermic and if the enthalpy change is positive, then the reaction is endothermic.