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:

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

So that's the enthalpy change.