Question

How do you calculate enthalpy change for H-H +I-I. 2H-IWhen H-H = 436kj/mol I-I=151 kj/mol H-I=297kj/mol

Answer 1

To calculate enthalpy change, what we do is to use the following formula:

`\Delta H^o_(Rxn)=\sum ^{}_{}(B.E,Reactants)-\sum ^{}_{}(B.E,Products)`

Where B.E represents the bonds energy.

Replacing, we got:

`\begin{gathered} \Delta H^o_(Rxn)=\sum ^{}_{}(B.E,Reactants)-\sum ^{}_{}(B.E,Products) \\ \Delta H^o_(Rxn)=(H-H)+(I-I)-2(H-I) \\ \Delta H^o_(Rxn)=436+151-2(297) \\ \Delta H^o_(Rxn)=-(7KJ)/(mol) \end{gathered}`

So that's the enthalpy change.