The following reaction has a standard enthalpy of reaction of -146.0 kJ/mol:
Cu2O(s) + 1/2 O2(g) -> 2CuO(s)
The standard enthalpy of formation of Cu2O(s) is -168.6 kJ/mol. What is the standard enthalpy of formation of CuO(s)?
To determine the standard enthalpy of formation of CuO(s), we need to use the given enthalpy values and apply the concept of Hess's Law. Hess's Law states that the overall enthalpy change in a reaction is independent of the pathway taken and depends only on the initial and final states of the system.
Let's assume the standard enthalpy of formation of CuO(s) is ΔHf.
The given reaction equation is:
Cu2O(s) + 1/2 O2(g) -> 2CuO(s)
We can break down this reaction into two steps:
Step 1: Formation of Cu2O(s) from its elements (Cu and O2):
Cu(s) + 1/2 O2(g) -> Cu2O(s) ΔH1
Step 2: Formation of CuO(s) from Cu2O(s):
Cu2O(s) -> 2CuO(s) ΔH2
According to Hess's Law, the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the individual steps.
ΔHoverall = ΔH1 + ΔH2
Considering that the enthalpy change is an extensive property, we can multiply the ΔH2 by a factor of 2:
ΔHoverall = ΔH1 + 2ΔH2
Given that the enthalpy change of the overall reaction (ΔHoverall) is -146.0 kJ/mol and the standard enthalpy of formation of Cu2O(s) (ΔH1) is -168.6 kJ/mol, we can rewrite the equation as:
-146.0 kJ/mol = -168.6 kJ/mol + 2ΔH2
Solving for ΔH2:
ΔH2 = (-146.0 kJ/mol + 168.6 kJ/mol) / 2
= 11.3 kJ/mol
Thus, the standard enthalpy of formation of CuO(s) is 11.3 kJ/mol.