NO2(g) → 1⁄2N2(g) + O2(g) ΔH1 = -33.9 kJ
NO2(g) → NO(g) + 1⁄2O2(g) ΔH2 = 56.5 kJ
What is ΔH for the reaction 1⁄2N2(g) + 1⁄2O2(g) → NO(g) ?
To find ΔH for the reaction 1⁄2N2(g) + 1⁄2O2(g) → NO(g), we can use the concept of Hess's Law.
Hess's Law states that if a chemical reaction can be expressed as the sum of two or more reactions, then the enthalpy change for the overall reaction is the sum of the enthalpy changes of the individual reactions.
In this case, we can see that the given reaction can be obtained by adding the two given reactions:
1⁄2 * (NO2(g) → 1⁄2N2(g) + O2(g)) + (NO2(g) → NO(g) + 1⁄2O2(g))
Now, to use Hess's Law, we need to manipulate the given reactions to ensure that the coefficients of the species we want to cancel out are the same:
1⁄2 * (NO2(g) → 1⁄2N2(g) + O2(g)) becomes:
1⁄4 * NO2(g) → 1⁄4 * N2(g) + 1⁄2 * O2(g)
We can now sum the two reactions:
1⁄4 * NO2(g) → 1⁄4 * N2(g) + 1⁄2 * O2(g) + (NO2(g) → NO(g) + 1⁄2O2(g))
Cancelling out the common species on both sides of the equation, we get:
1⁄4 * NO2(g) → 1⁄4 * N2(g) + 1⁄2 * O2(g) + NO(g)
The enthalpy change for the overall reaction is then the sum of the enthalpy changes for the individual reactions:
ΔH = ΔH1 + ΔH2
Substituting the given values:
ΔH = -33.9 kJ + 56.5 kJ
ΔH = 22.6 kJ
Therefore, the ΔH for the reaction 1⁄2N2(g) + 1⁄2O2(g) → NO(g) is 22.6 kJ.