Given the reaction: NaOH(s) + H_2O(l) -> Na^"+"(aq) + OH^"-" (aq) + 10.6 kJNaOH(s)+H2O(l)Na+(aq)+OH-(aq)+10.6kJ, the heat of reaction, ΔH, is (positive/negative), the entropy, ΔS, is (positive/negative) and the reaction is (spontaneous/not spontaneous). Thermochem?

1 Answer
May 26, 2017

DeltaH is negative
DeltaS is positive
DeltaG must be negative and thus reaction is spontaneous

Explanation:

NaOH(s)+H_2O(l)->Na(aq)+OH^"-" (aq) +10.6kJ

This is a dissolving reaction (not balanced).

In this reaction, heat is released (the amount of heat that is released is 10.6kJ). Therefore this reaction is exothermal. An exothermal reaction is a reaction in which the enthalpy of the system is lowered, therefore the DeltaH<0.
This may seem weird, because of the heat we observe coming from the reaction, but that heat is released from the reacting compounds and therefore not part anymore of the system.

For the entropy, it is important to note that this reaction is a dissolving reaction. The NaOH (s) is dissolved in water. This changes the state of the NaOH from (s) to (aq) and therefore, the amount of disorder is increased and thus DeltaS>0

To determine whether the reaction is spontaneous, we must look at the Gibbs free energy:

DeltaG=DeltaH-TxxDeltaS

We know now that DeltaH is negative and DeltaS is positive. Let's give H the value x and S the value y. Both x and y are positive. We obtain:

DeltaG=-x-Txxy
T is the temperature in Kelvin and can therefore not be lower than 0. If we fill in random positive numbers for x and y and T, we can only obtain negative values of DeltaG.

DeltaG<0 color(blue)("Spontaneous")
DeltaG=0 color(blue)("Reversible")
DeltaG>0 color(blue)("Not spontaneous")

Therefore the reaction will be spontaneous.