Observe the following reaction: ABO_{3(s)} xr | vedclass

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(A) The relationship between enthalpy change and internal energy change is given by: $\Delta_{r} H = \Delta_{r} U + \Delta n_{g} RT$. Given: $\Delta_{

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$x - 8.3$

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(A) The relationship between enthalpy change and internal energy change is given by: $\Delta_{r} H = \Delta_{r} U + \Delta n_{g} RT$. Given: $\Delta_{r} H = x \ kJ \ mol^{-1}$,$T = 1000 \ K$,$R = 8.3 \ J \ mol^{-1} \ K^{-1} = 8.3 	imes 10^{-3} \ kJ \ mol^{-1} \ K^{-1}$. For the reaction $ABO_{3(s)} ightarrow AO_{(s)} + BO_{2(g)}$,the change in the number of gaseous moles is $\Delta n_{g} = (n_{g})_{products} - (n_{g})_{reactants} = 1 - 0 = 1$. Substituting the values: $x = \Delta_{r} U + (1 	imes 8.3 	imes 10^{-3} \ kJ \ mol^{-1} \ K^{-1} 	imes 1000 \ K)$. $x = \Delta_{r} U + 8.3$. Therefore,$\Delta_{r} U = x - 8.3 \ kJ \ mol^{-1}$.

Observe the following reaction: $ABO_{3(s)} \xrightarrow{1000 \ K} AO_{(s)} + BO_{2(g)}$. $\Delta_{r} H$ for this reaction is $x \ kJ \ mol^{-1}$. What is its $\Delta_{r} U$ (in $kJ \ mol^{-1}$) at the same temperature? $(R = 8.3 \ J \ mol^{-1} \ K^{-1})$

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