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 ΔH FORMATION

H2(g) + F2(g) ---->2 HF(g) ΔH = -537 kJ

C(s) + 2 F2(g)----> CF4(g) ΔH = -680 kJ

2 C(s) + 2 H2(g)------> C2H4(g) ΔH = +52.3 kJ

Calculate ΔH for the reaction of ethylene with F2:

C2H4(g) + 6 F2(g) ----> 2 CF4(g) + 4 HF(g) ΔHRXN = ????

2.Use Hess’s Law to calculate ΔH for the reaction

NO(g) + O(g) ---> NO2(g) given the following information:

NO(g) + O3(g) ---> NO2(g) + O2(g) ΔH = -198.9 kJ

O3(g) ----> 1.5 O2(g) ΔH = -142.3 kJ

O2(g) ---> 2 O(g) ΔH = +495.0 kJ

3.Use Hess’s Law to calculate ΔH for the reaction N2O(g) + NO2(g) ----> 3 NO(g) given the following information:

2 NO(g) + O2(g) ----> 2 NO2(g) ΔH = -113.1 kJ

N2(g) + O2(g) ----> 2 NO(g) ΔH = +180.7 kJ

2 N2O(g) -----> 2 N2(g) + O2(g) ΔH = -163.2 kJ

Answer 1

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