Final answer:
The entropy change in the surroundings is -0.524 kJ/K. The sign of the entropy change for the system is positive, indicating an increase in entropy due to the formation of gaseous products. The sign for the entropy change of the universe is negative but close to zero, indicating that the reaction is non-spontaneous.
Step-by-step explanation:
(a) To calculate the entropy change in the surroundings with this reaction at 25°C, we can use the equation ΔSsurr = -ΔHrxn / T. Plugging in the values, we get: ΔSsurr = -(+163.2 kJ) / (25+273) K = -0.524 kJ/K.
(b) The sign of the entropy change for the system can be determined based on the ΔSrxn value. If ΔSrxn > 0, the reaction is endothermic and the entropy of the system increases. If ΔSrxn < 0, the reaction is exothermic and the entropy of the system decreases. In this case, the reaction is endothermic (+163.2 kJ) and the entropy of the system increases due to the formation of gaseous products.
(c) The sign for the entropy change for the universe can be determined based on the ΔSsurr and ΔSsys values. If ΔSsurr + ΔSsys > 0, the entropy of the universe increases and the reaction is spontaneous. If ΔSsurr + ΔSsys < 0, the entropy of the universe decreases and the reaction is non-spontaneous. In this case, since ΔSsurr is negative (-0.524 kJ/K) and ΔSsys is positive, the entropy change for the universe is negative but close to zero, indicating that the reaction is non-spontaneous.
The entropy change in the surroundings is negative, due to an endothermic reaction. The sign of the entropy change for the system is negative because the number of gas molecules decreases. Since both the system and surroundings have a negative entropy change, the total entropy change for the universe is negative, indicating that the reaction is nonspontaneous at 25°C.
The student's question is about calculating the entropy change in the surroundings for the reaction between nitrogen and oxygen gas to form dinitrogen monoxide at 25°C, determining the sign of the entropy change for the system, and for the universe, as well as discussing the spontaneity of the reaction.
Entropy Change in the Surroundings
To calculate the entropy change in the surroundings (ΔSsurroundings) for an exothermic or endothermic reaction at a constant temperature, we use the following formula:
ΔSsurroundings = -ΔHrxn / T
For this reaction, ΔHrxn is +163.2 kJ, and T is 25°C or 298 K. Therefore, the calculation is ΔSsurroundings = -(+163.2×1000 J)/(298 K) leading to ΔSsurroundings to be negative, indicating an increase in disorder in the surroundings.
Entropy Change for the System
The sign of the entropy change for the system (ΔSsystem) can be determined by looking at the states and number of molecules of the reactants and products. Here, there is a decrease in the number of gas molecules from 3 moles on the reactant side to 2 moles on the product side, which implies a negative ΔSsystem.
Entropy Change for the Universe
The total entropy change for the universe (ΔSuniverse) is the sum of ΔSsystem and ΔSsurroundings. Since both are negative, ΔSuniverse is also negative, and the reaction is nonspontaneous under standard conditions at 25°C.