Question

For combustion of one mole of magnesium in an open container at 300K and 1bar pressure, Δ_CH^Θ=–601.70kJ mol^–1, the magnitude of change in internal energy for the reaction is______ kJ. (Nearest integer) (Given : R = 8.3 J K^–1 mol^–1)

Answer 1

Correct Answer: 600

To solve this problem, we need to find the change in internal energy (ΔU) for the combustion of magnesium. The relationship between the change in enthalpy (ΔH) and the change in internal energy (ΔU) is given by the equation: ΔU = ΔH - ΔnRT. Here, Δn is the change in the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin.

Step 1: Identify Δn
For the combustion of magnesium: Mg(s) + 1/2 O₂(g) → MgO(s).
The initial moles of gas: 0.5 (from 1/2 O₂).
The final moles of gas: 0 (as MgO is solid).
Δn = moles of products - moles of reactants = 0 - 0.5 = -0.5.

Step 2: Calculate ΔU
Given: Δ_CH^Θ = -601.70 kJ/mol, R = 8.3 J K^-1 mol^-1, T = 300 K.
Convert R to kJ: R = 8.3 × 10^-3 kJ K^-1 mol^-1.
Substitute into the equation: ΔU = ΔH - ΔnRT.
ΔU = -601.70 kJ/mol - (-0.5)(8.3 × 10^-3 kJ K^-1 mol^-1)(300 K).
ΔU = -601.70 kJ/mol + 1.245 kJ/mol.
ΔU = -600.455 kJ/mol.

Step 3: Round to Nearest Integer
The magnitude is approximately 600.455 kJ/mol, so rounding gives 600 kJ/mol.

Validation
Ensure the computed value falls within the provided range of 600, 600. As our final result is exactly 600, it is within the expected range.

Conclusion
The magnitude of the change in internal energy is 600 kJ/mol.