Enthalpy-Entropy-Free energy

Chemistry Level 2

The above is a model for an exothermic reaction in which all reactants and products are gas state. What are the signs of Δ H \Delta H , Δ S \Delta S and Δ G \Delta G for this reaction?

Δ H \Delta H (-), Δ S \Delta S (-), Δ G \Delta G (-) Δ H \Delta H (-), Δ S \Delta S (+), Δ G \Delta G (-) Δ H \Delta H (+), Δ S \Delta S (+), Δ G \Delta G (-) Δ H \Delta H (-), Δ S \Delta S (+), Δ G \Delta G (+)

Lakshan Dumpa by Lakshan Dumpa , 24, India

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4 solutions

Mj Bltz
Oct 27, 2014

Since the reaction is stated to be exothermic, then Δ H \Delta H is negative. Since the number of gas molecules increased, from 6 to 9,then Δ S \Delta S is positive. Since Δ H < 0 \Delta H < 0 and Δ S > 0 \Delta S > 0 , then from the equation Δ G = Δ H − T Δ S \Delta G = \Delta H - T\Delta S , Δ G \Delta G is deduced to be negative.

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Enthalpy: a thermodynamic quantity equivalent to the total heat content of a system. It is equal to the internal energy of the system plus the product of pressure and volume. the change in enthalpy associated with a particular chemical process.

Entropy: a thermodynamic quantity representing the unavailability of a system's thermal energy for conversion into mechanical work, often interpreted as the degree of disorder or randomness in the system.

Free energy: a thermodynamic quantity equivalent to the capacity of a system to do work.

Verification:

Exothermic means release of heat that the chemicals become less energy that heat change is negative.

Potential to produce mechanical work decreased mean a decreased of quality of energy, that disorder increased. Total number of molecules increased supports disorder of a system being increased. Entropy increased as disorder increased.

Capacity to do work decreased when potential of reactive becomes idle. Therefore, free energy decreased.

Answer: Δ H ( − ) , Δ S ( + ) , Δ G ( − ) \boxed{\Delta H (-), \Delta S(+), \Delta G(-)}

Lu Chee Ket - 5 years, 4 months ago
Nitin Kumar
Oct 29, 2014

enthalpy is -ve , spontinity is +ve gibbs energy is -ve

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Fang Woei
May 14, 2014

Involve Bond breaking>> exothermic reaction >> dH -ve

no. of molecules increase >> degree of disorder/randomness increase >> dS +ve

the reaction is occur (spontaneous) >> dG -ve

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bond breaking is endothermic and not exothermic!

Mayank Holmes - 7 years ago
Shantanu Lankr
May 13, 2014

Since, the reaction is exothermis, the heat is rejected to the surroundings. Therefore, dH and dG will be negative. However, the entropy of the system will increase therefore dS will be positive.

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