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(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by the equation: $\Delta H = \Delta E + \De

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$H_{2_{(g)}} + Br_{2_{(g)}} \rightarrow 2HBr_{(g)}$

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(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by the equation: $\Delta H = \Delta E + \Delta n_{g} RT$.For $\Delta H = \Delta E$ to hold true,the change in the number of moles of gaseous species $(\Delta n_{g})$ must be $0$.Calculating $\Delta n_{g}$ for option $B$: $H_{2_{(g)}} + Br_{2_{(g)}} ightarrow 2HBr_{(g)}$.$\Delta n_{g} = (	ext{moles of gaseous products}) - (	ext{moles of gaseous reactants}) = 2 - (1 + 1) = 0$.Since $\Delta n_{g} = 0$,it follows that $\Delta H = \Delta E$.

Assume each reaction is carried out in an open container. For which reaction will $\Delta H = \Delta E$ ?

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