For a particular reaction, ?H = -99.84 kJ and ?S = -16.80 J/K. Calculate ?G for this reaction at 298 K. Number kJ What can be said about the spontaneity of the reaction at 298 K? The system is spontaneous as written. The system is at equilibrium. The system is spontaneous in the reverse direction.
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Since the final answer is required in kJ, we will convert ΔS to kJ/K. ΔS = -16.80 J/K * (1 kJ / 1000 J) = -0.0168 kJ/K Now, we can use the formula for Gibbs free energy (ΔG) at a given temperature (T): ΔG = ΔH - TΔS where ΔH is the enthalpy change, T is the Show more…
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For a certain chemical reaction, $\Delta H^{\circ}=-35.4 \mathrm{kJ}$ and $\Delta S^{\circ}=-85.5 \mathrm{J} / \mathrm{K}$ . (a) Is the reaction exothermic or endothermic? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the system?(c) Calculate $\Delta G^{\circ}$ for the reaction at 298 $\mathrm{K}$ . (d) Is the reaction spontaneous at 298 $\mathrm{K}$ under standard conditions?
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