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:

$\Delta S_{surr}=-\frac{Q}{T}$

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:

$\Delta S_{surr}=-\frac{163.2kJ}{298K}$

$\Delta S_{surr}=-\frac{163.2\times 1000J}{298K}$

$\Delta S_{surr}=-547.6J/K$

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