Calculate the temperature (in ºC) to which silver ore, Ag2O, must be heated to spontaneously decompose the ore to oxygen gas and solid silver metal at standard state. (ΔHºf of Ag2O(s) = –31.05 kJ/mol; Sº values: Ag(s), 42.55 J/mol•K; O2(g), 205.1 J/mol•K; Ag2O(s), 121.3 J/mol•K)
A. 374 ºC
B. 234 ºC
C. 467 ºC
D. 195 ºC
E. 843 ºC
2Ag2O ==> 4Ag + O2
dGo = dHo-TdSo
dHo is the negative of dHf which makes it 2*31.05 = 62.1 kJ
dSorxn = (4*dSo Ag + dSo O2)-(2*dSo Ag2O)
Substitute dHo and dSo into the dGo equation, make dGo = 0 which it will be at equilibrium, solve for T (remember T iws in kelvin), then subtract 273 to obtain T in C. Post your work if you get stuck.
To calculate the temperature at which the Silver oxide (Ag2O) will spontaneously decompose to oxygen gas (O2) and solid silver (Ag) at standard state, we can use the relationship between entropy change and Gibbs free energy change.
The standard free energy change (ΔGº) for the reaction can be calculated using the equation:
ΔGº = ΔHº - TΔSº
Where:
ΔGº is the standard free energy change
ΔHº is the standard enthalpy change
T is the temperature in Kelvin
ΔSº is the standard entropy change
In this case, we want to find the temperature at which ΔGº becomes zero, indicating a spontaneous reaction. Thus, we can set ΔGº = 0 and solve for T.
0 = ΔHº - TΔSº
Rearranging the equation:
TΔSº = ΔHº
Simplifying:
T = ΔHº / ΔSº
Now we can substitute the given values:
ΔHºf (Ag2O) = -31.05 kJ/mol
ΔSº (Ag) = 42.55 J/mol•K
ΔSº (O2) = 205.1 J/mol•K
ΔSº (Ag2O) = 121.3 J/mol•K
We need to convert the enthalpy change from kJ/mol to J/mol:
ΔHº = -31.05 kJ/mol * 1000 J/ 1 kJ = -31,050 J/mol
Substituting the values into the equation:
T = (-31,050 J/mol) / (121.3 J/mol•K)
T ≈ -256 K
Since temperature cannot be negative, it means the reaction will not occur spontaneously at any temperature.
Therefore, the correct answer is None of the above.