2 H2O (l) → 2 H2 (g) + O2 (g)
Using the thermodynamic values to calculate accurate values of ΔHr and ΔSr, will the reaction be product favored at high temperature? Why?
Thank you
To determine if the reaction is product-favored at high temperatures, we need to compare the entropy change (ΔS) and enthalpy change (ΔH) of the reaction.
First, let's calculate the ΔH value for the reaction using thermodynamic values:
ΔH = (2ΔHf[H2(g)] + ΔHf[O2(g)]) - (2ΔHf[H2O(l)])
Where:
ΔHf[H2(g)] = standard enthalpy of formation of H2(g)
ΔHf[O2(g)] = standard enthalpy of formation of O2(g)
ΔHf[H2O(l)] = standard enthalpy of formation of H2O(l)
Next, let's calculate the ΔS value for the reaction using thermodynamic values:
ΔS = (2ΔSf[H2(g)] + ΔSf[O2(g)]) - (2ΔSf[H2O(l)])
Where:
ΔSf[H2(g)] = standard entropy of formation of H2(g)
ΔSf[O2(g)] = standard entropy of formation of O2(g)
ΔSf[H2O(l)] = standard entropy of formation of H2O(l)
Now, if ΔHr (enthalpy change) is negative and ΔSr (entropy change) is positive, the reaction is exothermic (gives off heat) and favored in the forward direction (product-favored) at low temperatures.
However, at high temperatures, we need to consider both the enthalpy and entropy changes. If ΔHr is negative and ΔSr is positive, the reaction is still exothermic and favored in the forward direction. But, at high temperatures, the effect of entropy becomes even more prominent.
Typically, at high temperatures, the increase in temperature promotes the formation of more gaseous products due to the increase in entropy. In this case, the formation of gaseous H2 and O2 from liquid H2O will be favored at high temperatures. Therefore, the reaction would be product-favored at high temperatures.
It's important to note that the accuracy of the values in the thermodynamic calculations and the exact temperature range should be considered for a more precise conclusion.