Calculate the enthalpy change for the reaction NO(g)+ O(g) $\to$ NO₂(g) From the following data:

Question

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Using Hess's Law, we can calculate the enthalpy change for the reaction by breaking it down into a series of steps, for which we have known enthalpy changes:

Step 1: 2NO(g) + O2(g) -> 2NO2(g)     ΔH = -114.1 kJ
Step 2: 2NO2(g) + H2O(l) -> HNO2(aq) + HNO3(aq)     ΔH = -77.15 kJ
Step 3: HNO2(aq) + HNO3(aq) -> N2O5(g) + H2O(l)     ΔH = -40.3 kJ

Adding these three steps gives us the overall reaction:

2NO(g) + O2(g) + H2O(l) -> N2O5(g) + H2O(l)      ΔH = -231.6 kJ

However, the given reaction is not quite the same as the above reaction. We need to reverse the first step and change the coefficients of the second and third steps to match the reaction we want:

-2NO2(g) -> -2NO(g) - O2(g)      ΔH = 114.1 kJ
2HNO3(aq) -> -2NO(g) + 2H2O(l) + 3O2(g)     ΔH = 154.3 kJ
N2O5(g) + H2O(l) -> 2HNO3(aq)   ΔH = -95.1 kJ

Adding these three steps gives us the overall reaction we want:

NO(g) + O2(g) -> NO2(g)     ΔH = -114.1 kJ
2NO(g) + O2(g) -> 2NO2(g)    ΔH = -114.1 kJ x 2 = -228.2 kJ
2NO2(g) -> N2O5(g) + H2O(l)      ΔH = -95.1 kJ

Adding these gives us the overall enthalpy change for the reaction:

ΔH = -228.2 kJ + (-95.1 kJ) = -323.3 kJ

However, we need to account for the fact that the reaction we want has one NO molecule and one O2 molecule as reactants, while the reaction we calculated involves two NO molecules and one O2 molecule. We can use the fact that the enthalpy change is proportional to the amount of reactants consumed to adjust the value:

ΔH = (-323.3 kJ / 2) x (1/2) = -80.8 kJ

Finally, we need to account for the fact that the given reaction involves NO and O atoms, rather than NO and O2 molecules. To obtain the enthalpy change for this reaction, we need to use the enthalpy change for the dissociation of O2(g) into atoms and the enthalpy change for the formation of O atoms from half of an O2 molecule:

O2(g) -> 2O(g)      ΔH = 495 kJ
O(g) + 0.5O2(g) -> O3(g)    ΔH = -142 kJ
0.5O3(g) -> O2(g)      ΔH = 74.2 kJ

Adding these gives us the enthalpy change for the formation of O atoms from O2:

ΔH = 495 kJ + (-142 kJ) + 74.2 kJ = 427.2 kJ

Thus, the enthalpy change for the reaction as given is:

ΔH = -80.8 kJ + 427.2 kJ = 346.4 kJ

However, we need to reverse the sign of this value because the reaction as given is exothermic (it releases heat):

ΔH = -(-346.4 kJ) = 346.4 kJ

Rounded to one decimal place, this gives us a final answer of -304.1 kJ.

Calculate the Enthalpy Change for the Reaction NO(g)+ O(g) →\to→ NO2(g) from the Following Data: A)-551

Calculate the Enthalpy Change for the Reaction
NO(g)+ O(g) $\to$ NO₂(g)
from the Following Data:

A)-551