Calculate the DHof for the following reaction under standard state conditions:
2 Ag2S(s) + 2 H2O(l) ---> 4 Ag(s) + 2 H2S(g) + O2(g)
Given the following information:
Ag2S(s) ÄHof = -32.6 kJ/mol
H2O(l) Ä Hof = -285.83 kJ/mol
H2S(g) ÄHof = -20.6 kJ/mol
To calculate the standard enthalpy change (ΔHof) for a reaction, you need to use the Hess's Law approach. Hess's Law states that the overall enthalpy change for a reaction is equal to the sum of the individual enthalpy changes for each step of the reaction.
In this case, there are two steps involved. The first step is the decomposition of Ag2S(s) and H2O(l) to form Ag(s) and H2S(g). The second step is the formation of O2(g).
Step 1: Decomposition of Ag2S(s) and H2O(l)
Ag2S(s) + H2O(l) → 2 Ag(s) + H2S(g)
The given enthalpy changes for Ag2S, H2O, and H2S are:
ΔHof(Ag2S) = -32.6 kJ/mol
ΔHof(H2O) = -285.83 kJ/mol
ΔHof(H2S) = -20.6 kJ/mol
Since the balanced equation shows 2 moles of Ag2S(s) and H2O(l), we need to multiply the enthalpy change of each reactant by its stoichiometric coefficient:
2 * ΔHof(Ag2S) + 2 * ΔHof(H2O) → 4 * ΔHof(Ag) + 2 * ΔHof(H2S)
Substituting the given values:
2 * (-32.6 kJ/mol) + 2 * (-285.83 kJ/mol) → 4 * ΔHof(Ag) + 2 * (-20.6 kJ/mol)
Step 2: Formation of O2(g)
Since O2(g) is an element in its standard state, its enthalpy change will be zero: ΔHof(O2) = 0 kJ/mol.
Therefore, the overall equation becomes:
2 * (-32.6 kJ/mol) + 2 * (-285.83 kJ/mol) = 4 * ΔHof(Ag) + 2 * (-20.6 kJ/mol) + 0
Simplifying the equation:
-65.2 kJ/mol - 571.66 kJ/mol = 4 * ΔHof(Ag) - 41.2 kJ/mol
Rearranging the equation:
4 * ΔHof(Ag) = -65.2 kJ/mol - 571.66 kJ/mol + 41.2 kJ/mol
Calculating the right side:
4 * ΔHof(Ag) = -595.66 kJ/mol
Finally, solving for ΔHof(Ag):
ΔHof(Ag) = (-595.66 kJ/mol) / 4
Calculating the value:
ΔHof(Ag) = -148.92 kJ/mol
Therefore, the standard enthalpy change (ΔHof) for the given reaction is -148.92 kJ/mol under standard state conditions.