For the following exothermic reaction, predict under which conditions the reaction will be spontaneous. 2Mg(s) + O2(g) → 2MgO(s)
a. The reaction is always spontaneous.
b. The reaction is spontaneous at high temperatures.
c. The reaction is never spontaneous.
d. Insufficient data is provided to answer this question.
e. The reaction is spontaneous at low temperatures.

e. The reaction is spontaneous at low temperatures.

Explanation:

The reaction 2Mg(s) + O2(g) → 2MgO(s)  is an exothermic reaction. Let us recall that in an exothermic reaction, heat and light are given out and ΔH is negative (less than zero).

In an exothermic reaction increase in temperature shifts the equilibrium position towards the left and more reactants are produced. However, decrease in temperature shifts the equilibrium position towards the right and more products are obtained.

Also, since the reaction is exothermic, then given that; ΔG = ΔH - TΔS, ΔH is negative for an exothermic reaction and ΔG must be negative for a reaction to be spontaneous. Hence, at low temperature, the value of ΔH - TΔS remains negative (less than zero) hence the reaction is spontaneous.

Hence, Being an exothermic reaction, the process is spontaneous at low temperatures.