To determine the internal energy change (ΔU) for the reaction at $300 \mathrm{~K}$, we will use the relationship between enthalpy change (ΔH) and internal energy change:

$$\Delta H = \Delta U + \Delta nRT$$

where $\Delta H$ is the enthalpy change, $\Delta U$ is the internal energy change, $\Delta n$ is the change in moles of gas, $R$ is the universal gas constant, and $T$ is the temperature.

Given:

• $$\Delta H = +15 \mathrm{~kJ} = 15000 \mathrm{~J}$$
• $T = 300 \mathrm{~K}$
• $$R = 8.314 \, \mathrm{J \, mol^{-1} \, K^{-1}}$$
• $$\Delta n = \text{moles of products} - \text{moles of reactants}$$

From the balanced chemical equation:

$$\mathrm{A}_2(\mathrm{~g}) + 3 \mathrm{B}_2(\mathrm{~g}) \rightarrow 2 \mathrm{AB}_3(\mathrm{~g})$$

Moles of reactants = 1 (for $\mathrm{A}_2$) + 3 (for $\mathrm{B}_2$) = 4 moles

Moles of products = 2 (for $\mathrm{AB}_3$)

Therefore, $\Delta n$ = 2 (products) - 4 (reactants) = -2

Now, substituting these values into the equation:

$$\Delta H = \Delta U + \Delta nRT$$

$$15000 = \Delta U + (-2)(8.314)(300)$$

Simplify the equation:

$$15000 = \Delta U - 4988.4$$

Therefore:

$$\Delta U = 15000 + 4988.4 = 19988.4 \mathrm{~J}$$

Hence, the internal energy change is:

Option A: $19988.4 \mathrm{~J}$