Question

In: Chemistry

Be sure to answer all parts Take all the thermodynamic data from your textbook. Assume the...

Be sure to answer all parts

Take all the thermodynamic data from your textbook. Assume the data do not vary with temperature.

A. Calculate Delta S univ for the reaction H2O2 (l) ----> H2O2(g) at 179.0 C

B. Is the reaction spontaneous, nonspontaneous, or an equilibrium process?

C. The reaction NH3 (g) + HCl(g) ----> NH4Cl(s) is spontaneous in the forward direction at room temperature. Determine the temperature at which it is no longer spontaneous in the forward direction

D. Determine the boiling point of Br2

Expert Solution

A. For a reaction

ΔSreaction=ΔnpSproducts−ΔnpSreactants

from the termodynamic data in you rteexbook you substract the S of the gas state (g) from the liquid state (l).

the data that i´ve found is that for (g) = 232.7 J/molK and or (l) = 109.6 J/molK.

ΔSreaction = 232.7 - 109.6

ΔSreaction = 123.1 J/Kmol.

B. When using the entropy change of a process to assess spontaneity, it is important to carefully consider the definition of the system and surroundings. The second law of thermodynamics states that a process involving an isolated system will be spontaneous if the entropy of the system increases over time. In this case as we have a positive ΔS we can say that we have a spontaneous reaction, therefore the system releases energy when this change occurs.

C. The reaction becomes spontaneous when ΔG is negative. First find ΔS and then use that to find what temperature the reaction will become spontaneous by finding the temperature that ΔG will become negative.

ΔG=ΔH-TΔS here we find the T = ΔG - ΔH/ -ΔS assuming that we want our ΔG become 0 we arrange the equation

T= ΔH/ ΔS

calculate ΔH and ΔS from your table (i am using one table from a book)

1(-46.11) + 1(-92.3)] - [1(-314.43)] = 176.02 kJ/mol =ΔH

1(192.34) + 1(186.77)] - [1(94.56)] = 284.55 J/Kmol = ΔS

T = (176.02 kJ/mol) /(284.55 J/K mol)

T = 0.62 K = -272.53 °C the temperature will not longer be spontaneus above 0.62 K

D. ΔG = ΔH - ΔTS

At the boiling point, there is an equilibrium between liquid and gas phase:

0 = 30,910 - T (93.2)

T = 332 K

If you see any change in your textbook please change the data in the equation. Hope this works for you!

queen_honey_blossom answered 2 years ago

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