Question

Trevor dissolves sodium hydroxide pellets in a beaker of water at room temperature, and notes that the beaker becomes warm. Which correctly designates the signs of Delta. H, Delta. S, and Delta. G for this process? Delta. H > 0, Delta. S > 0, and Delta. G < 0 Delta. H < 0, Delta. S > 0, and Delta. G < 0 Delta. H > 0, Delta. S > 0, and Delta. G > 0 Delta. H < 0, Delta. S < 0, and Delta. G > 0.

Answer 1

The change in energy for the reaction is  $\Delta H$  < 0,    $\Delta S$ > 0, and $\Delta G$ < 0. Hence, option B is correct.

The chemical reaction with the loss of energy is exothermic reactions, and the reaction with the gain of energy is an endothermic reaction.

The change in enthalpy for the reaction is given by $\Delta H$, the change in entropy is given by $\Delta S$, and the change in free energy is given by $\Delta G$.

What is the energy change in the exothermic reaction?

The energy in the NaOH dissolution is released, which turns the solution warm.

• The release of the energy by the sample in dissolution makes the value  $\Delta H$ negative.

Thus, $\Delta H$  is < 0 for the reaction.

• The release of the energy results in increased disorder in the reaction mixture. The value  $\Delta S$ becomes positive.

Thus, the  $\Delta S$ is > 0 for the reaction.

• The reaction for the exothermic process is spontaneous. Thus, the value  $\Delta G$ is negative for the reaction.

Thus, the $\Delta G$ < 0.

Hence. the change in energy for the reaction is  $\Delta H$  < 0,    $\Delta S$ > 0, and $\Delta G$ < 0. Hence, option B is correct.

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