Question

Calculate the entropy change in the surroundings associated with this reaction occurring at 25∘c.

Answer 1

This is an incomplete question, here is a complete question.

Consider the reaction between nitrogen and oxygen gas from dinitrogen monoxide.

`2N_2(g)+O_2(g)\rightarrow 2N_2O(g)`

Given: delta Hrxn= +163.2kJ

Calculate the entropy change in the surroundings when this reaction occurs at 25 degrees C.

Answer : The entropy change in the surroundings is, -547.6 J/K

Explanation :

Formula used to calculate the entropy change in the surroundings is:

`\Delta S_(surr)=-(Q)/(T)`

where,

`\Delta S_(surr)` = entropy change in the surrounding

Q = heat energy

T = temperature =
`25^oC=273+25=298K`

Given:

`\Delta H_(rxn)=+163.2kJ`

`\Delta H_(rxn)=Q=+163.2kJ`

Now put all the given values in the above formula, we get:

`\Delta S_(surr)=-(163.2kJ)/(298K)`

`\Delta S_(surr)=-(163.2* 1000J)/(298K)`

`\Delta S_(surr)=-547.6J/K`

Therefore, the entropy change in the surroundings is, -547.6 J/K