Question

Calculate deltaH for the reaction

NO (g) + O (g) ----> NO2(g)

given following
NO + O3 ----> NO2 + O2 deltaH=-198.9 kj
O3 ---> 3/2 O2 deltaH= -142 kj
O2 -----> O2(g) deltaH=495

Answer 1

No one answer it?

Answer 2

Respuesta 304.1kj

Answer 3

From the given informations you have:

NO + O -> NO2 ∂H?
1. NO + O3 -> NO2 + O2 ∂H = -198.9 KJ
2. O3 -> 3/2 O2 ∂H = -142.3 KJ
3. O2 -> 2O ∂H = 495,0 KJ

First of all you want to manipulate the equations in order to match with the target equation (NO + O -> NO2)

So you get:

1. NO + O3 -> NO2 + O2 ∂H = -198.9 KJ (NO and NO2 are respectively the reactant and the product as well as in the target equation, no manipulation needed);
2. 3/2 O2 -> O3 (This swap is needed to simplify the equation in the final step) ∂H = +142.3 KJ;
3. O -> 1/2 O2 ∂H = +247.5 KJ (Swapping reactant and product so 2O can match with O in the target equation, but since we have got 2O instead of one we want to divide everything by 2, including ∂H);

We now have:

NO + O3 -> NO2 + O2 ∂H = -198.9 KJ
3/2 O2 -> O3 ∂H = +142.3 KJ
O -> 1/2 O2 ∂H = +247.5 KJ

Hence NO + O -> NO2 ∂H will be 190.9 KJ
(Given from: -198.9 + 142.3 + 247.5 = 190.0)

To simplify the equation we want to take all products and all reactants together, thus we get:

NO + O3 + 3/2 O2 + O —> NO2 + O2 + 1/2 O2 + O3

1. O3 in products simplifies with O3 in reactants
2. 3/2(=1.5) O2 in products simplifies with O2 + 1/2 O2 = 1.5 O2 in reactants

Hence the remaining equation corresponds to the given one:

NO + O —> NO2