Hess's Law 1. State the Hess's Law.

Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Hess's Law
1.
State the Hess's Law.
2.
Calculate the enthalpy change for the reaction :
CAH4 (9) + H2 (9) - C2H6 (g)
Given the following data:
C2H4 (9) + 3 O2 (9) - 2 CO2 (9) + 2 H20 (1)
i)
ii)
2 C2H6 (9) + 7 0z (9) - 4 CO2 (9) + 6 H20 (1)
iii)
2 H2 (9) + O2 (g) - 2 H20 (U
AH° = - 1410.9 kJ
AH° = - 3119.4 k)
AH° = - 571.6 kJ
3.
Calculate the standard enthalpy change for the reaction,
C2H2 (9) + 2 H2 (9) - C2H6 (9)
Given that:
i)
2 C2H2 (9) + 5 O2 (9) - 4 CO2 (9) + 2 H20 (1)
ii)
2 C2H6 (9) + 7 02 (0) - 4 CO2 (9) + 6 H20 (U)
ii)
2 H2 (9) + O2 (9) - 2 H20 (()
AH° = - 2599 kJ
AH° = - 3119 kJ
AH° = - 572 kJ
4.
The enthalpy of combustion of methane is -891 kJmoli. Calculate the enthalpy of
formation of methane if the enthalpy of formation of carbon dioxide and water are -
394 kJmol and -286 kJmol respectively. Chemical equation for formation of
methane:
Cs) + 2H2(9) → CHa(9)
Given:
CHA(9) + 202(9) - 2H20U + CO2(9)
Cs) + Ozco) → CO2(9)
Hz(g) + V2 O2 - H20
AH = -891 kJmol!
AH = -394 kJmol!
AH = -286 kJmol!
TUTORIAL 2 CHM271 PRINCIPLES OF PHYSICAL CHEMISTRY
5.
Calculate the standard enthalpy of formation of liquid methanol, CH;OH(I), using the
following information:
C (graphite) + 2H2(g) + 1/2 O2(g) - CH;OH(I)
Given:
C (graphite) + O2 - CO:(9)
H2(g) + 2 O2 → H20(1)
CH:OH(I) + 3/2 O2(9) - CO2(9) + 2H20(1)
AH° = -393.5 kJ/mol
AH° = -285.8 kJ/mol
AH° = -726.4 kJ/mol
6.
Calculate AH for this reaction:
CHa(g) + NH3(g) - HCN(g) + 3H2(g)
Given:
N2(g) + 3 H2(g) -2 NH3(g)
C(s) + 2 Hz(g) – CHa(g)
H2(g) + 2 C(s) + N2(9) - 2 HCN(g)
AH = -91.8 kJ
AH = -74.9 kJ
AH = +270.3 kJ
Transcribed Image Text:Hess's Law 1. State the Hess's Law. 2. Calculate the enthalpy change for the reaction : CAH4 (9) + H2 (9) - C2H6 (g) Given the following data: C2H4 (9) + 3 O2 (9) - 2 CO2 (9) + 2 H20 (1) i) ii) 2 C2H6 (9) + 7 0z (9) - 4 CO2 (9) + 6 H20 (1) iii) 2 H2 (9) + O2 (g) - 2 H20 (U AH° = - 1410.9 kJ AH° = - 3119.4 k) AH° = - 571.6 kJ 3. Calculate the standard enthalpy change for the reaction, C2H2 (9) + 2 H2 (9) - C2H6 (9) Given that: i) 2 C2H2 (9) + 5 O2 (9) - 4 CO2 (9) + 2 H20 (1) ii) 2 C2H6 (9) + 7 02 (0) - 4 CO2 (9) + 6 H20 (U) ii) 2 H2 (9) + O2 (9) - 2 H20 (() AH° = - 2599 kJ AH° = - 3119 kJ AH° = - 572 kJ 4. The enthalpy of combustion of methane is -891 kJmoli. Calculate the enthalpy of formation of methane if the enthalpy of formation of carbon dioxide and water are - 394 kJmol and -286 kJmol respectively. Chemical equation for formation of methane: Cs) + 2H2(9) → CHa(9) Given: CHA(9) + 202(9) - 2H20U + CO2(9) Cs) + Ozco) → CO2(9) Hz(g) + V2 O2 - H20 AH = -891 kJmol! AH = -394 kJmol! AH = -286 kJmol! TUTORIAL 2 CHM271 PRINCIPLES OF PHYSICAL CHEMISTRY 5. Calculate the standard enthalpy of formation of liquid methanol, CH;OH(I), using the following information: C (graphite) + 2H2(g) + 1/2 O2(g) - CH;OH(I) Given: C (graphite) + O2 - CO:(9) H2(g) + 2 O2 → H20(1) CH:OH(I) + 3/2 O2(9) - CO2(9) + 2H20(1) AH° = -393.5 kJ/mol AH° = -285.8 kJ/mol AH° = -726.4 kJ/mol 6. Calculate AH for this reaction: CHa(g) + NH3(g) - HCN(g) + 3H2(g) Given: N2(g) + 3 H2(g) -2 NH3(g) C(s) + 2 Hz(g) – CHa(g) H2(g) + 2 C(s) + N2(9) - 2 HCN(g) AH = -91.8 kJ AH = -74.9 kJ AH = +270.3 kJ
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