Use the bond enthalpies listed below to estimate the enthalpy change for the reaction
`H_(2)(g)+Br_(2)(g)rarr2HBr(g)`
Given:
`BE` of `H_(2), Br_(2)`, and `HBr` is `435, 192`, and `372 kJ mol^(-1)`, respectively.

Explanation

To estimate the enthalpy change for the reaction:

\[ \text{H}_2(g) + \text{Br}_2(g) \rightarrow 2 \text{HBr}(g) \]

we will use the bond enthalpies provided for the bonds involved in the reaction. The bond enthalpies given are:

- Bond enthalpy of \(\text{H}_2\) = 435 kJ/mol
- Bond enthalpy of \(\text{Br}_2\) = 192 kJ/mol
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