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 =?

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

A. ∆S ° (reaction) = -113 J / mol K

B. ∆S ° (reaction) = 125.8 J / mol K

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

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

∆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  

∆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

Delta H solution  

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an exothermic reaction  

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as endothermic or exothermic  

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first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j