Calculating K from Free Energy Equation

Question

The reaction 2H(g) + O(g) → 2HO(g) is used to fuel some liquid-propellant rockets. Although it has an unfavorable change in entropy (ΔS = −88.84 J/K), it is very exothermic, with a negative enthalpy of reaction (ΔH = −438.6 kJ). Below what temperature (in K) will this reaction be spontaneous? (Combustion reactions in rocket engines can reach around 3500 K.)

Answer 1

If a reaction is spontaneous, then its change in Gibbs Free Energy is less than 0 (i.e.  ΔG < 0).

The Gibbs Free Energy equation is  ΔG = ΔH - TΔS.

So, since  ΔG < 0,  then ΔH - TΔS < 0.

We are given ΔH = −438.6 kJ and ΔS = −88.84 J/K. Notice the inconsistent energy units (kJ vs. J) which is typical of this type of problem. So, convert −88.84 J to kJ:

−88.84 J * 1 kJ/1,000J = −0.08884 kJ

So, ΔS = −0.08884 kJ/K.

Now, plug what we know into ΔH - TΔS <0:

−438.6 kJ - T(−0.08884 kJ/K) <0

−438.6 kJ + T(0.08884 kJ/K) <0

T(0.08884 kJ/K) < 438.6 kJ

T(0.08884/K) < 438.6

divide both sides by 0.08884/K to isolate T --> T < 4937 K.

So, the reaction is spontaneous at temperatures below 4937 K.

(Since combustion reactions in rocket engines can reach around 3500 K, which is less than 4937 K, this reaction will be spontaneous for that usage).