Hess’s Law states that the Enthalpies of reaction are additive. It is possible t
ID: 635658 • Letter: H
Question
Hess’s Law states that the Enthalpies of reaction are additive. It is possible to use Hess’s law to calculate the enthalpy of change for a reaction if you know enthalpies of reaction of related reactions, bond energies or the standard heats of formation of the reactants and products.
Model 1:
Reaction:
Enthalpy: ??Hrxn in kJ (kilojoules)
H2(g) + ½ O2(g) ? H2O (l)
- 285.8 kJ
2 H2(g) + O2(g) ? 2 H2O (l)
- 571.6 kJ
H2O (l) ? H2(g) + ½ O2(g)
+ 285.8 kJ
C (s) + ½ O2(g) ? CO (g)
- 110.5 kJ
2 C (s) + O2(g) ? 2 CO (g)
- 221.0 kJ
CO (g) ? C (s) + ½ O2(g)
+ 110.5 kJ
C (s) + O2(g) ? CO2(g)
- 393.5 kJ
CO (g) + ½ O2(g) ? CO2(g)
- 283.0 kJ
H2O (l) + CO (g) ? H2 (g) + CO2 (g)
+ 2.8 kJ
1)What is the difference between reaction 1 and reaction 2 in the model? Answer:
2)What would have to be done to reaction 4 to make it reaction 5? Answer:
3)When the coefficients of a reaction are multiplied by a factor, what happens to the enthalpy of the reaction? Answer:
4)What was done to reaction 1 to make it reaction 3 or reaction 4 to make it reaction 6? Answer:
5)What is the relationship between the enthalpy of reaction for reaction 1 and 3 or 4 and 6? Answer:
6)When the reaction is reversed, where the products are written as the reactants, what happens to the enthalpy of the reaction? Answer:
Reaction:
Enthalpy: ??Hrxn in kJ (kilojoules)
H2(g) + ½ O2(g) ? H2O (l)
- 285.8 kJ
2 H2(g) + O2(g) ? 2 H2O (l)
- 571.6 kJ
H2O (l) ? H2(g) + ½ O2(g)
+ 285.8 kJ
C (s) + ½ O2(g) ? CO (g)
- 110.5 kJ
2 C (s) + O2(g) ? 2 CO (g)
- 221.0 kJ
CO (g) ? C (s) + ½ O2(g)
+ 110.5 kJ
C (s) + O2(g) ? CO2(g)
- 393.5 kJ
CO (g) + ½ O2(g) ? CO2(g)
- 283.0 kJ
H2O (l) + CO (g) ? H2 (g) + CO2 (g)
+ 2.8 kJ
Explanation / Answer
1) In reaction 1 , only 1 mole of H2O is formed whereas in reaction 2 , there is production of 2 moles of H2O I.e reaction 1 is multiplied by factor 2 to get reaction 2.
2) again reaction 4 is multiplied by a factor of 2 to make it reaction 5 .
3) when the coefficients are multiplied by a factor , then the reaction enthalpy is also multiplied by the same factor . Because enthalpy of formation of any compound is written for 1 mole so whenever it is different than 1 mole , then enthalpy changes by same factor.
4)Reaction 1 is reversed to make it reaction 3 i.e the products in reaction 1 are written as reactants in reactants in reaction 3 .
Similarly reaction 4 is reversed to get reaction 6. I.e if we see clearly reaction 6 is nothing but reverse form or backward reaction of reaction 1.
5) because reaction 1 and 3 , 4 and 6 are reverse of each other so their enthalpy of reaction are equal in magnitude and opposite in sign I.e if forward reaction has negative enthaly then it's reverse reaction must have positive enthalpy and vice versa.
6) when we reverse the reaction , it's sign gets changed I.e endothermic becomes exothermic and vice versa.