What is the change in enthalpy of the first reaction below, given the enthalpies of the other two reactions?
2NO(g) + O2(g) + 2NO2(g)
N,(8)+ O2(g) → NO(g)
N,(8)+O2(g) + NO2(g)
AH = 90 kJ/mol
AH° = 34 kJ/mol
A. 56 kJ
B. -112 kJ
C. 124 kJ
D. -248 kJ

The correct option is;

B. -112 kJ

Explanation:

The parameters given are;

N (g) + O₂ (g) → NO (g) ΔH = 90 kJ/mol

N (g) + O₂ (g) → NO₂ (g) ΔH = 34 kJ/mol

The required chemical reaction is given as follows;

2NO (g) + O₂ (g) → 2NO₂ (g)

Therefore, the heat of formation of 2 moles of NO = 2 × 90 = 180 kJ

The heat of formation of 2 moles of NO₂ = 2 × 34 = 68 kJ

Hence, given that the heat of formation of O₂ at room temperature = 0 kJ/mol, we have;

Change in enthalpy of the chemical reaction = Heat of formation of the products - Heat of formation of the reactants

Change in enthalpy of the chemical reaction = 68 kJ - 180 kJ = -112 kJ

Change in enthalpy of the chemical reaction = -112 kJ.

h let me think

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b