Transcript Entry Task: 05.12-13.10 Block #2
Entry Task: Oct. 1
st
Monday
Question: Define Hess’s Law (in your own words) You have 5 minutes!
Agenda:
• Discuss Hess’s Law (review from 1 st year) • In-class problems (see backside of notes) • HW: Pre-Lab Hess’s Law Lab
I can…
•
Manipulate thermochemical equations (Hess Law) to find enthalpy.
Hess’s Law
H is well known for many reactions, and it is inconvenient to measure H for every reaction in which we are interested.
However, we can estimate H using published
H
values and the properties of enthalpy.
Hess’s Law
Hess’s law states that “[i]f a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”
Hess’s Law
Because H is a state function, the total enthalpy change depends only on the initial state of the reactants and the final state of the products.
Sample Exercise 5.8
Using Three Equations with Hess’s Law to Calculate
H
Calculate
H
for the reaction 2 C(
s
) + H 2 (
g
) C 2 H 2 (
g
) given the following chemical equations and their respective enthalpy changes: Flip Double keep 226.8 kJ
Sample Exercise 5.8
Using Three Equations with Hess’s Law to Calculate
H
Practice Exercise Calculate
H
for the reaction NO(
g
) + O(
g
) NO 2 (
g
) given the following information keep Flip Flip divide by 2
Answer:
–304.1 kJ
5.51
Desired: P
4
O
6
(g) + 2O
2
(g)
P
4
O
10
(s)
ΔH = ?
Flip keep P 4 (s) + 3O 2 P 4 (s) + 5O 2 (g) (g) P 4 O 6 (g) ΔH = -1640.1
P 4 O 10 (g) ΔH = -2940.1
-1300 kJ
5.53
Desired: C 2 H 4 (g) + 6F 2 (g) 2CF 4 (g) + 4HF (g) ΔH = ?
double double Flip H 2 (s) + F 2 (g) C(s) + 2F 2 (g) 2C(s) + 2H 2 (g) 2HF(g) ΔH = -537 CF 4 (g) ΔH = -680 C 2 H 4 (g) ΔH = +52.3
-2486.3 or -2.49x10
-3 kJ
Applying Hess law
With the following data:
a. S(s) + O 2 (g) b. 2SO 3 (g) 2SO SO 2 2 (g) (g) + O 2 (g) Use Hess law to calculate ΔH for
2S(s) + 3O 2 (g)
2SO 3 (g)
ΔH of -297 kJ ΔH of 198kJ
We want to find out: 2S(s) + 3O 2 (g) 2SO 3 (g) ΔH = ?
2S(s) + 2O 2 (g)
2SO 2
2SO 2 (g) + O 2 (g)
(g) ΔH of -594 kJ 2SO 3 (g) ΔH of -198kJ
2S(s) + 3O 2 (g)
2SO 3 (g) ΔH = -792 kJ
Applying Hess law
With the following data:
a. 2H 2 (g) + O 2 (g) Flip and double b. H 2 (g) + O 2 (g) 2H 2 O(l) ΔH= -527 kJ H 2 O 2 (l) ΔH= -188 kJ Use Hess law to calculate ΔH for
2H 2 O 2 (l)
2H 2 O(l) + O 2 (g)
We want to find out: 2H 2 O 2 (l) 2H 2 O(l) + O 2 (g) ΔH = ?
2H 2 O 2 (l) 2H 2 (g) + 2O 2 (g) 2H 2 (g) + O 2 (g) 2H 2 O(l) 376kJ -527kJ
2H 2 O 2 (l)
2H 2 O(l) + O 2 (g) ΔH = -151.0 kJ
Desired: 2CO(g) + 2NO (g) 2CO 2 (g) + N 2 (g) ΔH= ?
a. 2CO(g) + O 2 (g) b. N 2 (g) + O 2 (g) 2CO 2 (g) ΔH = -566.0 kJ 2NO(g) ΔH= 180.6kJ
a. 2CO(g) + O 2 (g) b. 2NO(g) O 2 (g) + N 2 2CO 2 (g) ΔH = -566.0 kJ (g) ΔH= -180.6kJ
2CO(g) + 2NO (g)
2CO 2 (g) + N 2 (g) ΔH= -746.6kJ
Desired: 4Al(s) + 3MnO 2 (s) 2Al 2 O 3 (s) + 3Mn(s) ΔH= ?
a. 4Al(s) + 3O 2 (g) b. Mn(s) + O 2 (g) 2Al 2 O 3 MnO 2 (s) ΔH = -3352 kJ (s) ΔH= -521 kJ a. 4Al(s) + 3O 2 (g) b. 3MnO 2 (s) 2Al 2 O 3 (s) ΔH = -3352 kJ 3Mn(s) + 3O 2 (g) ΔH= 1563 kJ
4Al(s) + 3MnO 2 (s)
2Al 2 O 3 (s) + 3Mn(s) ΔH= -1789kJ
H 2 O (g) + C (s) CO (g) + H 2 (g) ΔH= ?
a.
H 2 (g) + ½ O 2 (g) H 2 O (g) ΔH = - 242.0 kJ b.
2CO (g) 2 C (s) + O 2 (g) ΔH = + 221.0 kJ a. H 2 O(s) ½ O 2 (g) + H 2 (g) ΔH = 242.0kJ
b. C(s) + ½ O 2 (g) CO(g) ΔH = -110.5kJ
H
2
O (g) + C (s)
CO (g) + H
2
ΔH= 131.5kJ
(g)
2CO (g) + O 2 (g) 2CO 2 (g) =ΔH?
C (s) + O 2 (g) CO 2 (g) ΔH = - 393.5 kJ C (s) + ½ O 2 (g) CO (g) ΔH = - 110.5 kJ 2C (s) + 2O 2 (g) 2CO 2 (g) ΔH = - 787kJ 2CO(g) O 2 (g) + 2C (s) ΔH = 221kJ
2CO (g) + O
2
(g)
ΔH= -566kJ 2CO
2
(g)
C 2 H 4 (g) + H 2 (g) C 2 H 6 (g) =ΔH?
C 2 H 4 (g) + 3 O 2 (g) 2 CO 2 C H 2 2 H 6 (g) + 3½ O 2 (g) + ½ O 2 (g) 2 CO 2 (g) H 2 O (l) (g) + 2 H 2 O (l) ΔH -1411 kJ (g) + 3 H 2 O (l)ΔH = -1560 kJ ΔH = -285 kJ C 2 H 4 (g) + 3 O 2 (g) 2 CO 2 (g) + 2 H 2 O (l) ΔH -1411 kJ 2 CO 2 (g) + 3 H 2 O (l) C 2 H 6 (g) + 3½ O 2 (g) ΔH= 1560 kJ H 2 (g) + ½ O 2 (g) H 2 O (l) ΔH = -285. kJ C 2 H 4 (g) + H 2 (g) C 2 H 6 (g) ΔH= -136kJ
YOUR TASK TONIGHT