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calculate the change in the entropy of the surroundings and the change in the entropy of the universe for the freezing of 1.00 mole of water from the following data: Temperature = 225 K ∆H = -6030 J/mol ∆Ssystem = -22.1 J/molK



Answer :

According to the given statement The change in the entropy of the surroundings and the change in the entropy of the universe for the freezing 24.40 JK⁻¹.

Describe freezing and provide an example:

Freezing is the process through which a liquid solidifies. Freezing happens when heat is transferred from a substance since this atoms slow down and form tighter bonds. An example of freezing is when water turns into ice.

Briefing:

H₂(g)+12O₂(g)→H₂O(l),

ΔrH∘=−6030KJmol−1

ΔSsurr=qsurrT

Entropy change = S

T = Temperature in K [T = 225 K under ideal circumstances]

Heat absorbed by the environment is measured by qsurr (qsurr=+241.1KJmol1)

ΔSsurr=-6030 /247.1

ΔSsurr=24.40 JK⁻¹ .

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