Question

Calculate ΔH° = for the chemical reaction
Cl2(g) + F2(g) --->2ClF(g)

Bond Energies, kJ mol-1
F---F 159
Cl--Cl 243
Cl--F 255

Answer 1

Answer: $\Delta H^0$ for the reaction $Cl_2(g)+F_2(g)\rightarrow 2ClF(g)$  is -108kJ

Explanation:

The balanced chemical reaction is,

The expression for enthalpy change is,

$\Delta H=[n\times B.E_{reactants}]-[n\times B.E{products}]$

where, n = number of moles

Now put all the given values in this expression, we get

$\Delta H=[1\times B.E_{Cl_2}+1\times B.E_{F_2}]-[2\times B.E_{ClF}]$

$\Delta H=[(1\times 243)+(1\times 159)]-[2\times 255]$

$\Delta H=-108kJ$

Thus $\Delta H^0$ for the reaction $Cl_2(g)+F_2(g)\rightarrow 2ClF(g)$  is -108kJ