Please show some work For the reaction: NO(g) + 1/2 O2(g) → NO2(g) ΔH°rxn is -114.14 kJ/mol. Calculate ΔH°f of gaseous nitrogen monoxide, given that ΔH°f of NO2(g) is 33.90 kJ/mol. Answers: 181.9 kJ/mol -35.64 kJ/mol 91.04 kJ/mol 148.0 kJ/mol -114.1 kJ/mol

ΔH°f of gaseous nitrogen monoxide is 148.0 kJ/mol

Explanation:

The balanced chemical reaction is,

The expression for enthalpy change is,

where,

n = number of moles

(as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

Therefore, ΔH°f of gaseous nitrogen monoxide is 148.0 kJ/mol

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.

a