Achieve Week 5&6 #9

[andrea guzman 1F]

Consider the following reaction at 298 K
C(graphite) + 2H₂ (g) -> CH4 (g) delta H= -74.6 kJ
Whats the difference between solving change in entropy of the system, change in entropy of the surroundings, and change in entropy of the universe?

[Maleeha Amir 2E]

Change in entropy for the system is the sum of standard entropies for products minus the sum of standard entropies for reactants. Change in entropy for surrounding is (-delta H)/T of the system. For the universe its just the sum of the system and the surrounding

[906013563]

The change in entropy of the system during a reaction is determined by the entropy values of the reactants and products; you subtract the total entropy of the reactants from that of the products. For the surroundings, it's calculated from the heat the system exchanges with its surroundings, using the enthalpy change of the reaction divided by the temperature, with exothermic reactions increasing the entropy of the surroundings. The entropy change of the universe is the sum of the changes in the system and surroundings; in an exothermic process, the universe's entropy typically increases. Specifically, the given reaction is exothermic, releasing energy to the surroundings, which disperses as increased entropy. To calculate these entropy changes, you'd use tabulated standard molar entropies for the system and the reaction's enthalpy change for the surroundings.

Hope this helps!

[106046096]

Hi! For the change in entropy of the system, you would do the sum of entropies for the products minus the sum of entropies for the reactants. For the change in entropy of the surroundings, you would do the negative enthalpy of the system divided by the temperature. Lastly, the change in entropy of the universe would just be the change in entropy of the system plus the change in entropy of the surroundings.