For the reaction 3O_{2(g)} rightarrow 2O_{3(g | vedclass

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(C) 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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$RT$

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(C) 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 $3O_{2(g)} \rightarrow 2O_{3(g)}$,the change in the number of moles of gaseous species $(\Delta n_{(g)})$ is calculated as:$\Delta n_{(g)} = n_{p(g)} - n_{r(g)} = 2 - 3 = -1$.Substituting this value into the equation:$\Delta H = \Delta U + (-1)RT$$\Delta H = \Delta U - RT$.Rearranging the terms to find $\Delta U - \Delta H$:$\Delta U - \Delta H = RT$.

For the reaction $3O_{2(g)} \rightarrow 2O_{3(g)}$,the value of $\Delta U - \Delta H$ is equal to .....

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