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For a reaction, [tex]\(\Delta H = -286 \, \text{kJ}\)[/tex]. When is the reaction spontaneous?

A. When [tex]\(T\Delta S \ \textgreater \ 0\)[/tex]
B. When [tex]\(T\Delta S \ \textless \ -286 \, \text{kJ}\)[/tex]
C. When [tex]\(T\Delta S = -286 \, \text{kJ}\)[/tex]
D. When [tex]\(T\Delta S = 2(-286 \, \text{kJ})\)[/tex]


Sagot :

To determine when the reaction is spontaneous, we need to make use of the concept of Gibbs free energy change ([tex]\(\Delta G\)[/tex]). The formula for Gibbs free energy is:

[tex]\[ \Delta G = \Delta H - T \Delta S \][/tex]

where:
- [tex]\(\Delta G\)[/tex] is the change in Gibbs free energy.
- [tex]\(\Delta H\)[/tex] is the change in enthalpy.
- [tex]\(T\)[/tex] is the temperature in Kelvin.
- [tex]\(\Delta S\)[/tex] is the change in entropy.

For a reaction to be spontaneous, [tex]\(\Delta G\)[/tex] should be negative ([tex]\(\Delta G < 0\)[/tex]).

Given that [tex]\(\Delta H = -286 \, \text{kJ}\)[/tex], we want to determine the conditions under which [tex]\(\Delta G\)[/tex] is negative.

[tex]\[ \Delta G = -286 \, \text{kJ} - T \Delta S \][/tex]

For [tex]\(\Delta G\)[/tex] to be spontaneous ([tex]\(\Delta G < 0\)[/tex]):

[tex]\[ -286 \, \text{kJ} - T \Delta S < 0 \][/tex]

Add [tex]\(T \Delta S\)[/tex] to both sides of the inequality:

[tex]\[ -286 \, \text{kJ} < T \Delta S \][/tex]

Thus, to satisfy this condition:

[tex]\[ T \Delta S > 286 \, \text{kJ} \][/tex]

Therefore, the correct condition for spontaneity is:

[tex]\(T \Delta S > 286 \, \text{kJ}\)[/tex].

So, the correct answer is:

A. When [tex]\( T \Delta S > 0\)[/tex].
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