QNEET-2024-Ch-120: NEET 2024 Chemistry - Class 11 NEET-2024-Ch-120 Solution: For the following reaction at 300 K \(A_2(g) + 3B_2(g)…

To determine the internal energy change (_U) for the reaction at 300   K , we will use the relationship between enthalpy change (_H) and internal energy change: _ H = _ U + _ n R T where _ H is the enthalpy change, _ U is the internal energy change, _ n is the change in moles of gas, R is the universal gas constant, and T is the temperature. Given: _ H = + 15   kJ = 15000   J T = 300   K R = 8.314 J mol _ 1 K _ 1 _ n = moles of products _ moles of reactants From the balanced chemical equation: A 2 (   g ) + 3 B 2 (   g ) _ 2 AB 3 (   g ) Moles of reactants = 1 (for A 2 ) + 3 (for B 2 ) = 4 moles Moles of products = 2 (for AB 3 ) Therefore, _ n = 2 (products) - 4 (reactants) = -2 Now, substituting these values into the equation: _ H = _ U + _ n R T 15000 = _ U + ( _ 2 ) ( 8.314 ) ( 300 ) Simplify the equation: 15000 = _ U _ 4988.4 Therefore: _ U = 15000 + 4988.4 = 19988.4   J Hence, the internal energy change is: Option A: 19988.4   J