Hess’s Law Problem: Show your work neatly and methodically. 1. From the enthalpi
ID: 955331 • Letter: H
Question
Hess’s Law Problem: Show your work neatly and methodically.
1. From the enthalpies of reaction PUT *** BY ANY REACTIONS THAT CORRESPOND IN A H FORMATION
H2(g) + F2(g) ---->2 HF(g) H = -537 kJ
C(s) + 2 F2(g)----> CF4(g) H = -680 kJ
2 C(s) + 2 H2(g)------> C2H4(g) H = +52.3 kJ
Calculate H for the reaction of ethylene with F2:
C2H4(g) + 6 F2(g) ----> 2 CF4(g) + 4 HF(g) HRXN = ????
2.Use Hess’s Law to calculate H for the reaction
NO(g) + O(g) ---> NO2(g) given the following information:
NO(g) + O3(g) ---> NO2(g) + O2(g) H = -198.9 kJ
O3(g) ----> 1.5 O2(g) H = -142.3 kJ
O2(g) ---> 2 O(g) H = +495.0 kJ
3.Use Hess’s Law to calculate H for the reaction N2O(g) + NO2(g) ----> 3 NO(g) given the following information:
2 NO(g) + O2(g) ----> 2 NO2(g) H = -113.1 kJ
N2(g) + O2(g) ----> 2 NO(g) H = +180.7 kJ
2 N2O(g) -----> 2 N2(g) + O2(g) H = -163.2 kJ
Explanation / Answer
1)
consider the given reactions
a) H2 + F2 ---. 2HF
b) C + 2F2 ---> CF4
c) 2C + 2H2 ---> C2H4
d) C2H4 + 6F2 ---> 2CF4 + 4HF
we can see that
eq d = ( 2 x eq a ) + ( 2 x eq b) - ( eq c)
so
according to hess law
dH d = ( 2 x dH a ) + ( 2 x dH b) - ( dH c)
dH d = ( 2 x -537) + ( 2 x-680) - ( 52.3)
dH d = -2486.3 kJ
so
dH rxn for the reaction of ehtylene with F2 is -2486.3 kJ
2)
a) N0 + 03 --> NO2 + O2
b) O3 ---> 1.5 O2
c) O2 ----> 2O
d) NO + O ----> NO2
we can see that
d = a - b - ( c/2)
so
dH d = dH a - dHb - ( dHc / 2)
dH d = -198.9 + 142.3 - ( 495 / 2)
dH d = -304.1 kJ
so
dH for the reaction NO + O --> NO2 is -304.1 kJ
3)
a) 2 NO + O2 ---> 2 NO2
b) N2 + O2 ---> 2 NO
c) 2 N2O ----> 2N2 + O2
d) N2O + NO2 ---> 3 NO
we can see that
d = ( c/2) + b - ( a/2)
so
dHd = ( dHc /2) + b - ( dH a / 2)
dH d = ( -163.2 / 2) + 180.7 - ( -113.1 /2)
dH d = 155.65
so
dH for the reaction N2O + NO2 ---> 3 NO is 155.65 kJ