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Relation between enthalpy change and internal energy change
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For the following reaction at 300 K: $A_2(g)+3B_2(g)\rightarrow2AB_3(g)$, the enthalpy change is +15 kJ, then the internal energy change is:A 19988.4 JB 200 JC 1999 JD 1.9988 kJ$\Delta n_g=n_{products}-n_{reactants}=2-(1+3)=-2$. $\Delta H=\Delta U+\Delta n_gRT\Rightarrow15000=\Delta U-2\times8.314\times300\Rightarrow\Delta U=15000+4988.4=19988.4$ J.