Question

Consider the reaction: 4CO(g)+2NO2(g)?4CO2(g)+N2(g) Using the following information, determine ?H? for the reaction at 25?C. NO(g) ?H?f = +91.3 kJ/mol CO2(g) ?H?f = -393.5 kJ/mol 2NO(g)+O2(g)?2NO2(g) ?H? = -116.2 kJ/mol 2CO(g)+O2(g)?2CO2(g) ?H? = -566.0 kJ/mol Express your answer using one decimal place.

Answer 1

Look at the last equation: 2CO(g)+O2(g) -----> 2CO2 (g) delta H= -566

delta H = (heat of formation of products) - (heat of formation of reactants)
delta H = (2 x heat of formation of CO2) - (2 x heat of formation of CO + heat of formation of O2)
-566 kJ = (2 x -393.5 kJ) - (2 x heat of formation of CO) + 0 since the heat of formation of an element is 0
from this, heat of formation of CO = -111 kJ

Look at the third equation: 2NO(g)+O2(g)-----> 2NO2(g) delta H= -116.2

delta H = (2 x heat of formation of NO2) - (2 x heat of formation of NO + heat of formation of O2)
-116.2 kJ = (2 x heat of formation of NO2) - (2 x 91.3) + 0
from this, heat of formation of NO2 = 33.2 kJ

Now look at the given equation: 4CO(g)+2NO2(g) ---> 4CO2(g)+N2(g)

delta H = (4 x heat of formation of CO2) - (4 x heat of formation of CO) - (2 x heat of formation of NO2)
= (4 x -393.5 kJ) - (4 x -111 kJ) - (2 x 33.2 kJ) = -1196 kJ