studying for chem test and confused...?

Question

i have to calculate the standard entropy change (DeltaS^o) for the following equation:

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

and from the table in the book the values for the standard entropies are:

NH3(g) = 192.3 J/mol*K
O2(g) = 205.0
NO2(g) = 240.0
H2O(g) = 188.7

the answer in the back of the book for the problem is -112 J/K

no matter what i try i can't get the right answer :( does anyone know how to solve this?? thanks!!

Answer 1

To obtain the net entropy of the reaction we use the given formula:

Sum of Entropy of products-Sum of Entropy of reactants

Entropy of each Product/Reactant is: Respective stoichiometric coefficient times its Standard Entropy

In this case:

Products are Water and NO2:

Entropy of Water is (6*188.7): 1132.2 J/K

Entropy of NO2 is (4*240.0): 960 J/K
---------------------------------------------------
Reactants are O2 and NH3

Entropy of Oxygen is (7*205.0): 1435 J/K

Entropy of Ammonia is (4*192.3): 769.2 J/K

Hence Entropy of Reaction is Entropy of Products-Reactants

(1132.2+960)J/K - (1435+769.2)J/K = -112J/K

Answer 2

monkey,

The procedure is very simple. You first need to take the entropy values for each of these substances, and multiply them all by the coefficients found in the equation. For example, the total entropy of the ammonia in this reaction will be 4 x 192.3, since there are 4 molecules of NH3.

Next, add up the values for NH3 and O2 to get the total entropy of the reactants. Do the same to get the entropy of the products. Finally, take the total entropy of the products, and SUBTRACT the total entropy of the reactants. The result will be the entropy change of the reaction.

Answer 3

Try (SE=standard entropy)
4*SE(NO2)+ 6*SE(H2O)-7SE(O2)-4SE(NH3)
The only thing you should try is the correct approach.

Answer 4

hey.... iget that answer!!!!!!

deltaS= products-reactants
(4x240)+(6x188.7)- (4x192.3)+ (7x205)= -112