Assume each reaction is carried out in an ope | vedclass

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

Vedclass · Accepted Answer

$H_{2(g)} + Br_{2(g)} 	o 2HBr_{(g)}$

Vedclass · Answer

(A) The relationship between enthalpy change $(\Delta H)$ and internal energy change ($\Delta E$ or $\Delta U$) is given by the equation: $\Delta H = \Delta E + \Delta n_g RT$. Here,$\Delta n_g$ is the change in the number of moles of gaseous species,calculated as: $\Delta n_g = \sum n_g(	ext{products}) - \sum n_g(	ext{reactants})$. For $\Delta H = \Delta E$,the term $\Delta n_g RT$ must be equal to $0$,which implies $\Delta n_g = 0$. Evaluating the options: $A$: $H_{2(g)} + Br_{2(g)} 	o 2HBr_{(g)}$,$\Delta n_g = 2 - (1 + 1) = 0$. $B$: $C_{(s)} + 2H_2O_{(g)} 	o 2H_{2(g)} + CO_{2(g)}$,$\Delta n_g = (2 + 1) - 2 = 1$. $C$: $PCl_{5(g)} 	o PCl_{3(g)} + Cl_{2(g)}$,$\Delta n_g = (1 + 1) - 1 = 1$. $D$: $2CO_{(g)} + O_{2(g)} 	o 2CO_{2(g)}$,$\Delta n_g = 2 - (2 + 1) = -1$. Thus,the condition $\Delta H = \Delta E$ is satisfied for reaction $A$.

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

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