Assuming all reactions occur in an open vesse | vedclass

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

Vedclass · Accepted Answer

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

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(A) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$. For $\Delta H = \Delta U$,the term $\Delta n_g$ must be equal to $0$. $\Delta n_g$ is the difference between the sum of stoichiometric coefficients of gaseous products and gaseous reactants. For option $A$: $H_{2(g)} + Br_{2(g)} \rightarrow 2 HBr_{(g)}$,$\Delta n_g = 2 - (1 + 1) = 0$. Since $\Delta n_g = 0$,$\Delta H = \Delta U$ for this reaction.

Assuming all reactions occur in an open vessel,for which reaction will $\Delta H = \Delta U$?

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