Hess's Law1.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 (UAH° = - 1410.9 kJAH° = - 3119.4 k)AH° = - 571.6 kJ3.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 kJAH° = - 3119 kJAH° = - 572 kJ4.The enthalpy of combustion of methane is -891 kJmoli. Calculate the enthalpy offormation of methane if the enthalpy of formation of carbon dioxide and water are -394 kJmol and -286 kJmol respectively. Chemical equation for formation ofmethane:Cs) + 2H2(9) → CHa(9)Given:CHA(9) + 202(9) - 2H20U + CO2(9)Cs) + Ozco) → CO2(9)Hz(g) + V2 O2 - H20AH = -891 kJmol!AH = -394 kJmol!AH = -286 kJmol!TUTORIAL 2 CHM271 PRINCIPLES OF PHYSICAL CHEMISTRY5.Calculate the standard enthalpy of formation of liquid methanol, CH;OH(I), using thefollowing 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/molAH° = -285.8 kJ/molAH° = -726.4 kJ/mol6.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 kJAH = -74.9 kJAH = +270.3 kJ

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

Science

Chemistry

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

Chemistry: Principles and Practice

3rd Edition

ISBN:9780534420123

Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Chapter5: Thermochemistry

Section: Chapter Questions

Problem 5.79QE

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Concept explainers

Thermochemistry

Thermochemistry can be considered as a branch of thermodynamics that deals with the connections between warmth, work, and various types of energy, formed because of different synthetic and actual cycles. Thermochemistry describes the energy changes that occur as a result of reactions or chemical changes in a substance.

Exergonic Reaction

The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energy during the course of the reaction. So, exothermic reaction is one type of exergonic reaction. Exergonic reaction releases work energy in different forms like heat, light or sound. For example, a glow stick releases light making that an exergonic reaction and not an exothermic reaction since no heat is released. Even endothermic reactions at very high temperature are exergonic.

Question

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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