Is it possible to determine whether a process will occur naturally solely by examining the sign of \(\Delta H\) for the process? Explain.

Short Answer

Expert verified
No, it is not sufficient to examine just the sign of \(\Delta H\). Although a negative \(\Delta H\) generally contributes to spontaneity, the Second Law of Thermodynamics requires an increase in total entropy (\(\Delta S_{universe}\)). This depends also on the entropy change of the system (\(\Delta S_{system}\)), not just the enthalpy change (\(\Delta H\)). Both conditions must be met for a process to occur spontaneously.

Step by step solution

01

Understanding the \(\Delta H\) value

The \(\Delta H\) value corresponds to the change in enthalpy of a system. If \(\Delta H\) is negative, the system releases heat (exothermic process). If \(\Delta H\) is positive, the system absorbs heat (endothermic process).
02

Second Law of Thermodynamics

The Second Law of Thermodynamics states that for a process to occur spontaneously, the total entropy (disorder) of the universe (\( \Delta S_{universe} \)) must increase. Entropy change (\(\Delta S\)) in the system is related to \(\Delta H\) and temperature (T) by the equation \( \Delta S = \frac{\Delta H}{T} \) for reversible processes.
03

Considering both conditions for spontaneity

While a negative \(\Delta H\) (exothermic reaction) generally contributes to an increased \(\Delta S_{universe}\), it doesn't guarantee it: we also need to examine the entropy change of the system (\(\Delta S_{system}\)). For example, an exothermic process might be non-spontaneous if it leads to a drastic decrease in \(\Delta S_{system}\) (i.e., the system becomes more ordered), causing the \(\Delta S_{universe}\) to decrease.

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