Why delta s for surrounding for exothermic reaction is may be + or -?
As deltaS = -deltaH / T so in exothermic reaction delta H is negative . hence the value of delta S is positive.
Why delta s for surrounding for exothermic reaction is may be + or -?
Introduction:
In thermodynamics, the change in entropy (ΔS) is a measure of the disorder or randomness of a system. When considering the surrounding environment of an exothermic reaction, the change in entropy can be either positive or negative, depending on certain factors.
Factors influencing ΔS for surrounding:
1. Temperature: The temperature of the surrounding environment plays a crucial role in determining the sign of ΔS. If the temperature increases, there is a higher probability of increased disorder, resulting in a positive ΔS. Conversely, if the temperature decreases, the disorder tends to decrease, leading to a negative ΔS.
2. Heat transfer: Exothermic reactions release heat energy to the surrounding environment. The magnitude of this heat transfer affects the ΔS for the surrounding. If a significant amount of heat is transferred, it can increase the disorder and randomness of the surrounding, resulting in a positive ΔS. Conversely, a smaller heat transfer may not significantly affect the disorder, leading to a negative ΔS.
3. System size: The size or extent of the system undergoing the exothermic reaction also influences ΔS for the surrounding. If the system is relatively small, the amount of heat transferred to the surrounding may have a more significant impact on disorder, resulting in a higher chance of a positive ΔS. In contrast, if the system is large, the heat transfer may not have a substantial effect on the overall disorder, leading to a negative ΔS.
Conclusion:
The sign of ΔS for the surrounding in an exothermic reaction can be either positive or negative, depending on the temperature, heat transfer, and system size. It is important to consider these factors to determine the impact on the entropy change of the surrounding environment.
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