For the following reaction,the relation betwe | vedclass

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

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$\Delta H = \Delta U - RT$

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(B) 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 the reaction: $2 SO_{2(g)} + O_{2(g)} \rightarrow 2 SO_{3(g)}$.The change in the number of moles of gaseous species is calculated as: $\Delta n_g = \sum n_{p(g)} - \sum n_{r(g)}$.$\Delta n_g = 2 - (2 + 1) = 2 - 3 = -1$.Substituting this value into the equation,we get: $\Delta H = \Delta U + (-1) RT$.Therefore,$\Delta H = \Delta U - RT$.

For the following reaction,the relation between $\Delta H$ and $\Delta U$ is:
$2 SO_{2(g)} + O_{2(g)} \rightarrow 2 SO_{3(g)}$

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For the following reaction,the relation between $\Delta H$ and $\Delta U$ is:$2 SO_{2(g)} + O_{2(g)} \rightarrow 2 SO_{3(g)}$