Question

Use the data from this table of thermodynamic properties to calculate the values of ΔS°rxn for each of the following reactions at 25 °C.
A) 4NH3(g) + 7O2(g) ---> 4NO2(g) + 6H2O(g) Delta S rxn =?
B) 2H2O2(l) --> 2H2O(l) +O2 (g) Delta S rxn =?

Answer 1

Answer:

A) ΔSºrxn = - 112.5 J/molK

B) ΔSºrxn =- 120.48 J/molK

Explanation:

For an general equation aA + bB → cC + dD

ΔSºrxn = cSºC + dSºD -  aSºA - bSºB

Using a table of thermodynamic properties in J/molK like Petrucci 8th Ed

A) ΔSºrxn = (4*240,1) + (6*188,8) - (4*192,5)- (7*205,1)= - 112.5 J/molK

B) ΔSºrxn= (2*69,91) + (1*205,1) - (2*232,7)= -120.48 J/molK

Be careful with the states of matter

Answer 2

A. ∆S ° (reaction) = -113 J / mol K

B. ∆S ° (reaction) = 125.8 J / mol K

Further explanation

Entropy (with the symbol S) in thermodynamics indicated the degree of system disorder

Entropy like other thermodynamic terms can only be calculated from the initial and final changes

Entropy will also change in the process of changing energy forms

The value of ΔS ° can be calculated from standard entropy data

∆S ° (reaction) = ∑S ° (product) - ∑S ° (reagent)

From the CRC Handbook of Chemistry and Physics data table, 84th Edition (2004) we can find out the Standard thermodynamic Quantities for Chemical Substances at 25 ° C for entropy: (sources can be obtained from other data)

NH₃ (g) = 192.8 J / mol K

O₂ (g) = 205 J / mol K

H₂O (g) = 188.8 J / mol K

H₂O (l) = 70 J / mol K

NO₂ (g) = 240.1 J / mol K

H₂O₂ (l) = 109.6 J / mol K

• A) 4NH₃ (g) + 7O₂ (g) ---> 4NO₂ (g) + 6H₂O (g)  

∆S ° (reaction) = {4.S ° NO₂ (g) + 6.S ° H₂O (g)} - {4.S ° NH₃ (g) + 7.S ° O₂ (g)}  

∆S ° (reaction) = {4. 240.1 + 6,188.8} - {4. 192.8 + 7,205}  

∆S ° (reaction) = -113 J / mol K  

• B) 2H₂O₂ (l) -> 2H₂O (l) + O₂ (g)  

∆S ° (reaction) = {2.S ° H₂O (l) + 1.S ° O₂} - {2.S ° H₂O₂ (l)}  

∆S ° (reaction) = {2.70+ 1,205} - {2,109.6}  

∆S ° (reaction) = 125.8 J / mol K

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