# Is this reaction likely to be spontaneous? 2Li(s) + 2H2O(g) ----> 2LiOH(s) + H2(g) ΔH = -222 kJ

##### 1 Answer
Apr 1, 2015

Yes it would be spontaneous for temperatures less than about 5000 K

To determine if a reaction is spontaneous or not, one needs to consider the enthalpy $\Delta H$ and entropy $\Delta S$ values for the reaction.

The Gibbs free energy equation combines these two values along with temperature to produce a $\Delta G$ value. If $\Delta G$ is negative, the reaction is spontaneous as written (in the forward direction). The equation looks like this:

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

Temperature (in Kelvin) has an effect : this is summarized in the table below - DH is $\Delta H$ and DS is $\Delta S$ In this case you are not given $\Delta S$. However, the reaction is exothermic since $\Delta H$ is negative. The reaction is most likely spontaneous. You can stop here but if more detail is required, read on.

The technical answer depends on if the entropy is increasing in the reaction (a positive $\Delta S$) or decreasing (a negative $\Delta S$)

The $\Delta S$ value for a reaction = sum of entropies of products - sum of entropies of reactants. This can be calculated from molar entropy tables. Basically $\Delta S$ will be positive if entropy is increasing.

You can tell from many reactions just by looking if disorder (entropy) is increasing. In this case, it is not so clear so I did a calculation. The calculation looks like

$\Delta S =$$\sum {S}_{\text{products")-sumS_("reactants}}$

$\Delta S =$(S_(H_2(g))+S_(LiOH)(s))-(S_(Li)(s) + S_(H_2O)(g))

$\Delta S = \left(130.7 + 42.8\right) - \left(29.1 + 188.8\right) = - 44.4 J m o {l}^{- 1} {K}^{- 1}$
This is a decrease in entropy and unfavorable toward spontanaeity

Since both $\Delta S$ and $\Delta H$ are negative, this tells us the reaction is only spontaneous at low temperatures.

You are not given the temperature but the lithium is reacting with steam which is at minimum 373 K (100 degrees Celsius)

At this temperature
$\Delta G = \Delta H - T \Delta S = - 222 - \left(373 \cdot \frac{- 44.4}{1000}\right)$= -205.4

This is spontaneous. (the $\Delta S$ value was also converted to kJ)

Note as the temperature increases, the situation changes. At about 5000 K, the reaction would cease to be spontaneous