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 ΔH, ΔS, and ΔG for this process?
A. ΔH > 0, ΔS > 0, and ΔG < 0
B. ΔH < 0, ΔS > 0, and ΔG < 0
C. ΔH > 0, ΔS > 0, and ΔG > 0
D. ΔH < 0, ΔS < 0, and ΔG > 0

Answer 1

Alright, so this question covers the subject of spontaneous reactions. Reactions tend to be spontaneous if the products have a lower potential energy than the reactants or when the product molecules are less ordered than the reactant molecules. This may seem a little confusing but to put it simply, spontaneous reactions occur naturally. 
This means there is no external force for the reaction, usually exothermic, and increases entropy. Gibbs free energy change helps us determine if the reaction is spontaneous:

   delta G= delta H -T(delta S)       spontaneous if delta G<0

Remember that H stands for enthalpy, or potential energy; T is for temperature; and S is entropy, the amount of disorder(spontaneous reactions increase in entropy take for ice to liquid water).

- delta H= exothermic  
The reaction in the problem releases heat so the enthalpy is negative.

delta S increases with increased temperature
The entropy will be positive.

Plugging this in the Gibbs equation, you can assume delta G will be less than 0.

Therefore, the answer is B. Sorry for the long and possibly not helpful explanation. I'm learning this material currently myself. Best of luck!

Answer 2

Answer: B. ΔH < 0, ΔS > 0, and ΔG < 0

Explanation:

As the beaker gets warm ,the energy has been released when Trevor dissolves sodium hydroxide pellets in a beaker of water. Thus as the energy has been released in the reaction, the reaction is exothermic and the change in enthalpy i.e. $\DeltaH$ is negative and is less than zero.

The entropy is the measure of degree of randomness. The entropy increases when the randomness increases and the entropy decreases when the randomness decreases. When a substance dissolves in water, it dissociate into ions and hence the randomness increases thus the change in entropy i.e. $\DeltaS$ is positive and  is more than zero.

also $\Delta G=\Delta H-T\Delta S$

$\Delta G=(-ve)-T(+ve)=(-ve)(-ve)=-ve$

That is change in gibbs free energy i.e. $\Delta G$ is negative and is less than zero.