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A certain reaction has ΔH∘=35.4kJ and ΔS∘=85.0J/K. You may want to reference (Pages 831 - 832) Section 19.6 while completing this problem. Part APart complete Is the reaction exothermic or endothermic? exothermic endothermic Previous Answers Correct A positive enthalpy value indicates that energy is absorbed by the system. Therefore, the reaction is endothermic. Part BPart complete Does the reaction lead to an increase or decrease in the disorder of the system? increase decrease Previous Answers Correct The sign of the entropy change is positive, which indicates that the disorder, or entropy, of the system has increased. Part C Calculate ΔG∘ for the reaction at 298 K.

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Answer:

ΔG° for the reaction at 298 K is 10.070 kJ

Explanation:

Here we have ΔH° = 35.4 kJ

ΔS° = 85.0 J/K

The formula for ΔG° for the reaction is given by the following relation;

ΔG°  = ΔH° - T×ΔS°

Where:

ΔH° = Change in enthalpy of the chemical reaction

ΔS° = Change in entropy of the system and

ΔG° = Change in Gibbs free energy

T  = Temperature of reaction = 298 K

Therefore, ΔG°  = 35.4 kJ - 298 K × 85.0 J/K = 35.4 kJ - 298 K × 85.0 J/K

= 35.4 kJ - 25330 J = 35.4 kJ - 25.330 kJ = 10.070 kJ

Therefore, ΔG° for the reaction at 298 K = 10.070 kJ.

adioabiola

The change in Gibb's free energy, ΔG° for the reaction at 298 K is; 10.070 kJ

Given; we have;

  • ΔH° = 35.4 kJ

  • ΔS° = 85.0 J/K

We can evaluate ΔG° by using the formula below;

  • ΔG° = ΔH° - T×ΔS°

in which case;

  • ΔH° = Change in enthalpy of the chemical reaction = 35.4 kJ

  • ΔS° = Change in entropy of the system = 85.0J/K

  • ΔG° = Change in Gibb's free energy

  • T = Temperature of reaction = 298 K

Therefore,

  • ΔG° = 35.4 kJ - 298 K × 85.0 J/K

  • = 35.4 kJ - (298 K × 85.0 J/K)

  • = 35.4 kJ - 25330 J

  • = 35.4 kJ - 25.330 kJ

  • = 10.070 kJ

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