Question

determine the standard enthalpy of formation for 30 g of NO given the following information about the formation of NO2 under standard conditions, and DH (NO2)=+ 33.2 kJ/mol. DHrxn= -114.2 kJ

Answer 1

The balanced equation is written as

2NO(g) + O2(g) → 2NO2(g)

We know that DHrxn= sum of DHproducts-sum of DHreactants=2xDH(NO2,g)-[2xDH(NO,g)+DH(O2,g)]

Given DHrxn=-114.2 kJ, DH(NO2,g)=+33.2 kJ/mol, DH(O2,g)=0, DH(NO,g)=?

Given weight of NO=30 g, molecular weight of NO=30.01 g/mol,

moles of NO=weight/molecular weight=30 g/30 g/mol=1 mol.

From the above equation, mole ratio between NO and NO2=1:1, so moles of NO2=1 mol, and O2 =0.5 mol (NO:O2=2:1)

DHrxn=1molx2xDH(NO2, g)-[1molx2xDH(NO,g)+ 0.5 mol DH(O2,g)]

-114.2 kJ=1molx2x33.2 kJ/mol-[1molx2xDH(NO,g)+0.5 molx0]

DH(NO,g)=(180.6 kJ)/2 mol=90.3 kJ/mol

DH(NO,g)=90.3 kJ/mol.