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

answer: semi-infinite solid initially at a uniform temperature of 400 ◦c is cooled at its surface by convection. the cooling fluid at a constant temperature of 200 ◦c has a convective coefficient (ha) of 250 w/m2.k. the physical properties of the solid: k = 20 w/m.k and α = 4 * 10-5  m2/s. using δx = 0.04 m and δt = 10 s. write the finite difference equations describing the transient heat transfer of the first five δx and the temperature profile after 30 s.