For the process 2SO2(g) + O2(g) Right arrow. 2SO3(g), Delta. S = –187.9 J/K and Delta. H = –198.4 kJ at 297.0 K are known. What is the entropy of this reaction? Use Delta. G = Delta. H – TDelta. S.

The entropy of this reaction (ΔS) = -187.9 J/K

Explanation:

.Step : Data given

ΔS = -187.9 J/K

ΔH = -198.4 kJ

T = 297.0 K

Step 2: The balanced equation

2SO2(g) + O2(g) → 2SO3(g)

Step 3: Calculate the entropy

ΔG = ΔH - TΔS

⇒ΔG = the gibbs free energy

⇒ΔH = the change in enthalpy of the reaction = -198.4 kJ

⇒with T = the temperature = 297.0 K

⇒with ΔS = the change of entropy =  -187.9 J/K

ΔG = -198400 kJ -297*(-187.9kJ/mol)

ΔG = -198400+55806.3

ΔG = - 142593.7 kJ = the gibbs free energy

The entropy of this reaction (ΔS) = -187.9 J/K

metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

if it's not useful, feel free to report for mistake in answer.

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