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

No one answer it?

Respuesta 304.1kj

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

To calculate the enthalpy change (ΔH) for the given reaction, we will use the concept of Hess's Law. Hess's Law states that the overall enthalpy change for a reaction is the same regardless of the pathway taken.

The given reaction can be derived from the combination of two reactions:

1. Reaction 1: NO (g) + O3(g) → NO2(g) + O2(g) ΔH1 = -198.9 kJ
2. Reaction 2: O3(g) → 3/2 O2(g) ΔH2 = -142 kJ

Since we need to cancel out O3 in the final reaction, we need to reverse Reaction 2 and multiply it by 2/3 to balance the oxygen atoms:

3/2 O2(g) → O3(g) (reversed Reaction 2, multiplied by 2/3) ΔH2' = -142 kJ * (2/3) = -94.66 kJ

Now we can sum up the reactions to obtain the desired reaction:

NO (g) + O3(g) → NO2(g) + O2(g) (ΔH1)
3/2 O2(g) → O3(g) (ΔH2')

Adding the two reactions results in the desired reaction:

NO (g) + O3(g) + (3/2)O2(g) → NO2(g) + O2(g) + O3(g)

To calculate ΔH for the desired reaction, we add the enthalpy changes of the individual reactions:

ΔH = ΔH1 + ΔH2'
= -198.9 kJ + (-94.66 kJ)
= -293.56 kJ

Therefore, the ΔH for the reaction NO(g) + O(g) → NO2(g) is -293.56 kJ.