To calculate ΔH_rxn for the reaction 2NOCl(g) -> N2(g) + O2(g) + Cl(g), we can use the given set of reactions and their respective enthalpy changes (ΔH).
First, let's write the target reaction in terms of the given reactions:
2NOCl(g) -> 2(NO(g) + 1/2Cl2(g))
or
2NOCl(g) -> 2NO(g) + Cl2(g)
We can now sum up the reactions together and cancel out the common substances:
(1/2N2(g) + 1/2O2(g)) + (NO(g) + 1/2Cl2(g)) -> 2(NO(g) + 1/2Cl2(g)) + N2(g) + O2(g)
or
1/2N2(g) + 1/2O2(g) + NO(g) + 1/2Cl2(g) -> 2NO(g) + Cl2(g) + N2(g) + O2(g)
Now, let's calculate the overall change in enthalpy (ΔH_rxn) by summing up the enthalpy changes of the individual reactions:
ΔH_rxn = (1/2N2(g) + 1/2O2(g)) + (NO(g) + 1/2Cl2(g)) - (2NO(g) + Cl2(g) + N2(g) + O2(g))
Since the ΔH values are given as follows:
ΔH1 = 90.3 kJ (for 1/2N2(g) + 1/2O2(g) -> NO(g))
ΔH2 = -38.6 kJ (for NO(g) + 1/2Cl2(g) -> NOCl(g))
Substituting these values:
ΔH_rxn = (90.3 kJ) + (-38.6 kJ) - (2NO(g) + Cl2(g) + N2(g) + O2(g))
However, we still need the values for the enthalpies of formation of NO(g), Cl2(g), N2(g), and O2(g) to calculate the overall ΔH_rxn. Unfortunately, these values are not provided in the given information.
Hence, without the enthalpies of formation of the individual species, we cannot calculate the exact value of ΔH_rxn for the given reaction.