* Please provide solutions step by step very spcifically (If you provide solutio

Question

* Please provide solutions step by step very spcifically

    (If you provide solutions in handwriting, please write down very clearly. Thank you so much)

The enthalpy of vaporization of ammonia is 23.35 kJ/mol at its boiling point (–33.4°C). Calculate the value of DSsurr when 1.00 mole of ammonia is vaporized at –33.4°C and 1.00 atm.

A)

0

B)

–6.99 ´ 102 J/K mol

C)

9.74 ´ 101 J/K mol

D)

–9.74 ´ 101 J/K mol

E)

6.99 ´ 102 J/K mol

A)

0

B)

–6.99 ´ 102 J/K mol

C)

9.74 ´ 101 J/K mol

D)

–9.74 ´ 101 J/K mol

E)

6.99 ´ 102 J/K mol

Explanation / Answer

DSsystem +DSsurroundings =0

Generally for phase transition DSsystem=(enthalpy of vapourisation /Temperature )

So DSsystem=23350/(273-33.4)

97.4j/mol k

So DSsurr=-97.4j/mol k

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.