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If ΔH = 400 kjoule mole⁻¹ and ΔS = 0.2 kjoule K⁻¹ mole⁻¹ for a reaction 2x + y --> z, at which minimum temperature the reaction will be spontaneous? |
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Answer» 2000KExplanation:ΔH = 400 kjoule mole⁻¹ ΔS = 0.2 kjoule K⁻¹ For the REACTION at 298 K, the equation FORMED will be - 2A + B → C ∆H = 400 kJ mol–1 and ∆S = 0.2 kJ K–1 mol–1 Therefore, from the expression, ∆G = ∆H – T∆S Assuming the reaction at equilibrium, ∆T for the reaction would beT = ∆H - ∆G × 1/∆S= ∆H /∆S= 400/0.2kJ ( ∆G = 0 at equilibrium)T = 2000 K Thus, for the reaction to be spontaneous, ∆G must be negative. Hence, at a temperature of 2000 K the reaction will be spontaneous. |
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