Answer :
This is an incomplete question, here is a complete question.
Consider the reaction between nitrogen and oxygen gas from dinitrogen monoxide.
[tex]2N_2(g)+O_2(g)\rightarrow 2N_2O(g)[/tex]
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:
[tex]\Delta S_{surr}=-\frac{Q}{T}[/tex]
where,
[tex]\Delta S_{surr}[/tex] = entropy change in the surrounding
Q = heat energy
T = temperature = [tex]25^oC=273+25=298K[/tex]
Given:
[tex]\Delta H_{rxn}=+163.2kJ[/tex]
[tex]\Delta H_{rxn}=Q=+163.2kJ[/tex]
Now put all the given values in the above formula, we get:
[tex]\Delta S_{surr}=-\frac{163.2kJ}{298K}[/tex]
[tex]\Delta S_{surr}=-\frac{163.2\times 1000J}{298K}[/tex]
[tex]\Delta S_{surr}=-547.6J/K[/tex]
Therefore, the entropy change in the surroundings is, -547.6 J/K