Calculate the enthalpy change during the reaction `:`
`H_(2(g))+Br_(2(g))rarr 2HBr_((g))`
Given, `e_(H-H)=435kJ mol^(-1),e_(Br-Br)=192kJ mol^(-1)` and `e_(H-Br)=368kJ mol^(-1).`

Explanation

To calculate the enthalpy change during the reaction:

\[ H_{2(g)} + Br_{2(g)} \rightarrow 2HBr_{(g)} \]

we will use the bond enthalpies provided:

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