Given the following reaction: 2CO (?) + O2 (?) → 2CO2 (?) ∆? = −566 kJ/mol...

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Question

Given the following reaction: 2CO (?) + O2 (?) → 2CO2 (?) ∆? = −566 kJ/mol...

Given the following reaction:
2CO (?) + O2 (?) → 2CO2 (?) ∆? = −566 kJ/mol
Calculate the change in internal energy when ?∆? at constant pressure is -102 kJ/mol.

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Answer 1

Answer #1

By the definitions of thermodynammics, the enthalpy of a system is equal to the sum of its internal energy and the product of its pressure and volume.

i.e.

Differentiating both sides,

As it is mentioned that the system is in constant pressure, so P is a constant value and can be taken out of the differential.

Given :

= -566 kJ/mol

= -102 kJ/mol

The negative value of indicates that the reaction is exothermic, i.e. heat is given out during the reaction.

The negative value of indicates that the system is getting compressed. This is evident from the fact that 3 moles of gaseous reactants give rise to only 2 moles of gaseous products.

= -566 kJ/mol - ( -102 kJ/mol)

= -566 kJ/mol + 102 kJ/mol

= -464 kJ/mol

The negative value of indicates that the internal energy of the system has decreased. This is logical as a huge amount of energy has been given out of the system, although the compression had increased the internal energy of the system by a small amount.

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Answer 2

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