Question:

For combustion of one mole of magnesium in an open container at 300K and 1bar pressure, ΔCHΘ=–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)

Updated On: Sep 24, 2024
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Correct Answer: 600

Solution and Explanation

The correct answer is 600
\(Mg(s)+\frac{1}{2}O2→MgO(s)\)
\(ΔH=ΔU+ΔngRT\)
\(Δng=−\frac{1}{2}\)
\(−601.70=ΔU−\frac{1}{2}(8.3)(300)×10^{−3}\)
ΔU = –601.70 + 1.245
ΔU ≃ –600 kJ
Hence , Magnitude of change in internal energy is 600 kJ.

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Concepts Used:

Laws of Thermodynamics

Thermodynamics in physics is a branch that deals with heat, work and temperature, and their relation to energy, radiation and physical properties of matter.

The First Law of Thermodynamics:

The first law of thermodynamics, also known as the Law of Conservation of Energy, states that energy can neither be created nor destroyed; energy can only be transferred or changed from one form to another. 

The Second Law of Thermodynamics:

The second law of thermodynamics says that the entropy of any isolated system always increases. Isolated systems spontaneously evolve towards thermal equilibrium—the state of maximum entropy of the system. More simply put: the entropy of the universe (the ultimate isolated system) only increases and never decreases.

The Third Law of Thermodynamics:

The third law of thermodynamics states that the entropy of a system approaches a constant value as the temperature approaches absolute zero. The entropy of a system at absolute zero is typically zero, and in all cases is determined only by the number of different ground states it has. Specifically, the entropy of a pure crystalline substance (perfect order) at absolute zero temperature is zero