Question

Calculate ΔS°for the combustion of ammonia.

4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(l)

Substance NH3(g) O2(g) N2(g) H2O(l)
S°(J/K·mol) 192 205.1 192 70
-135 J

-579 J

-387 J

579 J

Answer 1

Answer: The
`\Delta S^o` of the reaction is
`-579JK^(-1)`

Step-by-step explanation:

Entropy change of the reaction is defined as the difference between the total entropy change of the products and the total entropy change of the reactants.

The equation representing entropy change of the reaction follows:

`\Delta S_(rxn)=\sum [n* \Delta S^o_(products)]-\sum [n* \Delta S^o_(reactants)]`

For the given chemical equation:

`4NH_3(g)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(l)`

We are given:

`\Delta S^o_(NH_3)=192Jmol^(-1)K^(-1)\\\Delta S^o_(O_2)=205.1Jmol^(-1)K^(-1)\\\Delta S^o_(N_2)=192Jmol^(-1)K^(-1)\\\Delta S^o_(H_2O)=70Jmol^(-1)K^(-1)`

Putting values in above equation, we get:

`\Delta S^o_(rxn)=[(6* \Delta S^o_(H_2O))+(2* \Delta S^o_(N_2))]-[(4* \Delta S^o_(NH_3))+(3* \Delta S^o_(O_2))]`

`\Delta S^o_(rxn)=[(6* 70)+(2* 192)]-[(4* 192)+(3* 205.1)]=-579JK^(-1)`

Hence, the
`\Delta S^o` of the reaction is
`-579JK^(-1)`