1) Consider the reaction: 2NH3(g) + 3N2O(g)4N2(g) + 3H2O(g) Using standard absol

Question

1) Consider the reaction:

2NH3(g) + 3N2O(g)4N2(g) + 3H2O(g)

Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.57 moles of NH3(g) react at standard conditions. S°system = J/K

2) Consider the reaction:

2N2(g) + O2(g)2N2O(g)

Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.71 moles of N2(g) react at standard conditions. S°system = J/K

3) Consider the reaction:

2BrF3(g)Br2(g) + 3F2(g)

Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.89 moles of BrF3(g) react at standard conditions. S°system = J/K

Explanation / Answer

1) First calculate ?S0rxn from standard absolute entropies

?S0rxn= ?S0 (products) - ?S0 (reactants)

= 4*S0(N2) + 3*S0(H2O) - [ 2*S0(NH3) +3*S0(N2O) ] ---------------[ as per the reaction given, the products are 4 moles of nitrogen & 3 moles of water and the reactants are 2 moles of ammonia and 3 moles of nitrous oxide]

= 4*191.6 + 3*188.8 - [ 2*192.77 + 3*219.6 ]

= 290 J/K

Now calculate the entropy change for 1.57 moles of NH3(g) reacting:

?S = 1.57 mol NH3(g) * [290 JK-1/1 mol NH3(g)]

= 455 J/K

Similar procedure can be followed for the other two parts as well.

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