• If we take the reaction between sodium and chorine, we know that this is a very exothermic reaction and also involves a decrease in entropy as a solid is produced from a solid and a gas (provided a flame is used to supply the necessary activation energy)

2Na (s) + Cl2?(g) → 2NaCl (s)

• If there is a decrease in entropy, how can the reaction be spontaneous?
• We need to take into account the entropy of the surroundings as well
  • The energy being released causes a substantial increase in entropy of the surroundings because there are more ways of arranging the quanta (packets of energy) in the surroundings than the system alone
• Therefore, the total entropy change for a reaction is

ΔSΘ?total?=?ΔS?Θsys?+ ΔSΘsurr

(sys = system and surr = surroundings)

• So, in the case of sodium and chlorine, the large amounts of energy released makes?ΔSΘsurr?very positive, which will outweigh the negative value of?ΔS?Θsys

Answer

ΔS?Θtotal?= ΔS?Θsys?+ ΔS?Θsurr

ΔSΘ?total?= -90.1 + 1379 =?1289 J K-1?mol-1

• Entropy changes are an order of magnitude smaller than enthalpy changes, so entropy is measured in joules rather than kilojoules. The full unit for entropy is?J K-1?mol-1
• The standard entropy change (ΔSΘsystem) for a given reaction can be calculated using the standard entropies (S??) of the reactants and products
• The equation to calculate the standard entropy change of a system is:

ΔSΘsystem?= ΣΔSΘproducts?- ΣΔSΘreactants

(where Σ = sum of)

• For example, the standard entropy change for the formation of ammonia (NH3) from nitrogen (N2) and hydrogen (H2) can be calculated using this equation

N2(g) + 3H2(g) ? 2NH3(g)

ΔSΘsystem?= (2 x ΔSΘ(NH3)) - (ΔSΘ(N2) + 3 x ΔSΘ(H2))

• Notice that, unlike enthalpy of formation for elements, entropy for elements is?not?zero and you can find entropy values for elements and compounds in data books

Answer

ΔSΘsystem?= ΣΔSΘproducts?- ΣΔSΘreactants

ΔSΘsystem?= (2 x 38.20) - (2 x 32.60 + 205.0)

=?-193.8 J K-1?mol-1

Answer:

ΔSΘsystem?= ΣΔSΘproducts?- ΣΔSΘreactants

ΔSΘsystem= [2 x?SΘ(NH3)] - [SΘ(N2)+ (3 x?SΘ(H2?))]

ΔSΘsystem= [2 x 192.3] - [191.6 + (3 x 131)]

ΔSΘsystem?= 384.6 - 584.6

ΔSΘsystem=?-200 J K-1?mol–1

• To calculate the entropy change of the surroundings,?ΔSΘ?surr?,?we need to know the energy that has been transferred to them
• This is given by the enthalpy change,?ΔH, and the relationship can be expressed as:

• T?is the absolute temperature
• The entropy change of the surroundings depends upon temperature
  • The transfer of a given quantity of energy to surroundings at a low temperature will produce a greater entropy change than the transfer of the same amount of energy to the surroundings at a higher temperature

Answer

• • 
  • Convert ΔH?from kJ mol-1?to J mol-1?by multiplying by 1000
  • 
  • = - 4483 J K-1 mol-1

转载自savemyexams