Part A: Calculate the enthalpy of the reaction 2NO(g)+O2(g)2NO2(g) given the fol
ID: 892624 • Letter: P
Question
Part A: Calculate the enthalpy of the reaction
2NO(g)+O2(g)2NO2(g)
given the following reactions and enthalpies of formation:
12N2(g)+O2(g)NO2(g), HA=33.2 kJ
12N2(g)+12O2(g)NO(g), HB=90.2 kJ
Part B: Calculate the enthalpy of the reaction
4B(s)+3O2(g)2B2O3(s)
given the following pertinent information:
B2O3(s)+3H2O(g)3O2(g)+B2H6(g), HA=+2035 kJ
2B(s)+3H2(g)B2H6(g), HB=+36 kJ
H2(g)+12O2(g)H2O(l), HC=285 kJ
H2O(l)H2O(g), HD=+44 kJ
Explanation / Answer
Part A
in the given data he has given two equation multiply both with 2 and add the both reaction
cancel the terms that are common in opposite side
N2(g) + 2O2(g) -----------> 2NO2(g) HA= 66.4 kJ -------1
N2(g) + O2(g) ------------> 2NO(g) HB=180.4 kJ
this can be written in the following way also
2NO(g) ------> N2(g) + O2(g) HB= -180.4 kJ ----2
now add eq 1 and 2
N2(g) + 2O2(g) -----------> 2NO2(g) HA= 66.4 kJ
2NO(g) ------> N2(g) + O2(g) HB= -180.4 kJ
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2NO(g) + O2(g) ------> 2NO2(g) = 66.4 - 180.4 = -114kJ
PartB
multiply eq 1 with 2 and reverse it
here multiply eq 2 with 2
multiply eq 3 with 6 and reverse it
multiply eq 4 with 6 and reverse it
now add all these and cancel with same substence which are in opposite side
6O2(g) + 2B2H6(g) ------------> 2B2O3 (s) + 6 H2O(g) HA = -4070 kJ
4B(s) + 6H2(g) -------------> 2B2H6(g) HB= +72 kJ
6H2O(l) -------------------------> 6H2(g) + 3O2(g) HC =+1710 kJ
6H2O(g) ------------------------> 6H2O(l) HD= -264 kJ
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4 B(s) + 3 O2(g) ------------------------> 2 B2O3(s) H = -4070 + 72 + 1710 - 264 = -2552 kJ