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For a certain reaction ΔH° = 219 kJ and ΔS° = –21 J/K. Determine whether the reaction is spontaneous or nonspontaneous. - Chemistry

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प्रश्न

For a certain reaction ΔH° = 219 kJ and ΔS° = –21 J/K. Determine whether the reaction is spontaneous or nonspontaneous.

थोडक्यात उत्तर
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उत्तर

Given:

∆H0 = 219kJ; ∆S0 = -21 J/K, ∆G0 = ?

For the standard state, T = 298 K

∆G0 = ∆H0 - T∆S0

= 219 - 298 × (-21) × 10-3

= 219 + 6.258

= 225.3 kJ

Since ∆S < 0 and ∆G0 > 0, the reaction is non-spontaneous.

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Spontaneous (Irreversible) Process
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पाठ 4: Chemical Thermodynamics - Exercises [पृष्ठ ८८]

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बालभारती Chemistry [English] Standard 12 Maharashtra State Board
पाठ 4 Chemical Thermodynamics
Exercises | Q 4.1 | पृष्ठ ८८

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