Respuesta :
Answer:
[tex]\Delta S_{surr} = + 0.32113\: kJ/K[/tex]
Explanation:
Given: Entropy of surrounding: ΔSsurr = ?
Temperature: T= 355 K
The change in enthalpy of reaction: ΔH = -114 kJ
Pressure: P = constant
As we know, ΔH = -114 kJ ⇒ negative
Therefore, the given reaction is an exothermic reaction
Therefore, Entropy of surrounding at constant pressure is given by,
[tex]\Delta S_{surr} = \frac{-\Delta H}{T}[/tex]
[tex]\therefore \Delta S_{surr} = -\left (\frac{-114 kJ}{355 K} \right ) = + 0.32113\: kJ/K > 0[/tex]
In the given reaction:
2NO(g) + O₂(g) → 2NO₂(g)
As, the number of moles of gaseous products is less than the number of moles of gaseous reactants.
[tex]\therefore \Delta S_{system} < 0[/tex]
As we know, for a spontaneous process, that the total entropy should be positive.
[tex]\Delta S_{total} = \Delta S_{surr} + \Delta S_{system} > 0[/tex]
Therefore, at the given temperature,
- if [tex]\Delta S_{surr} > \Delta S_{system} \Rightarrow \Delta S_{total} > 0[/tex] then the given reaction is spontaneous
- if [tex]\Delta S_{surr} < \Delta S_{system} \Rightarrow \Delta S_{total} < 0[/tex] then the given reaction is non-spontaneous
