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

May 26, 2017

$\Delta H$ is negative
$\Delta S$ is positive
$\Delta G$ must be negative and thus reaction is spontaneous

#### Explanation:

$N a O H \left(s\right) + {H}_{2} O \left(l\right) \to N a \left(a q\right) + O {H}^{\text{-}} \left(a q\right) + 10.6 k J$

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 $\Delta H < 0$.
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 $N a O H \left(s\right)$ is dissolved in water. This changes the state of the $N a O H$ from $\left(s\right)$ to $\left(a q\right)$ and therefore, the amount of disorder is increased and thus $\Delta S > 0$

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

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

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

$\Delta G = - x - T \times y$
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 $\Delta G$.

$\Delta G < 0$ $\textcolor{b l u e}{\text{Spontaneous}}$
$\Delta G = 0$ $\textcolor{b l u e}{\text{Reversible}}$
$\Delta G > 0$ $\textcolor{b l u e}{\text{Not spontaneous}}$

Therefore the reaction will be spontaneous.