Calculate the standard enthalpy of formation of CuO2, given that:
2Cu2O + O2 ==> 4 CuO (-288 kJ)
Cu2O ==> Cu + CuO (11 kJ)
PLEASE HELP!!!!
You need to examine your post. Nowhere in your post does CuO2 appear except in the question. Neither equation contains CuO2.
To calculate the standard enthalpy of formation (ΔHf) of CuO2, we need to use the given information and apply Hess's Law, which states that the overall enthalpy change of a reaction is independent of the route taken.
We are given the following reactions and their enthalpy changes:
1) 2Cu2O + O2 → 4CuO: ΔH = -288 kJ
2) Cu2O → Cu + CuO: ΔH = 11 kJ
The desired reaction is:
Cu + 1/2 O2 → CuO2
We can manipulate the given reactions to obtain the desired reaction by combining them in the following way:
2 × (2Cu2O + O2 → 4CuO) + 2Cu2O → 4Cu + 2CuO
Let's calculate the enthalpy change for this reaction:
2 × (-288 kJ) + 2 × (11 kJ) = -576 kJ + 22 kJ = -554 kJ
Therefore, the standard enthalpy of formation of CuO2 is -554 kJ.
To calculate the standard enthalpy of formation of CuO2, we can use a Hess's Law approach.
Hess's Law states that if a reaction can be expressed as the sum of two or more reactions, the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the individual reactions.
In this case, we can consider the formation of CuO2 as the combination of two reactions:
1. 2Cu2O + O2 → 4CuO
2. Cu2O → Cu + CuO
Given the enthalpy changes for these two reactions, we can calculate the standard enthalpy change for the overall reaction:
The first reaction has a standard enthalpy change of -288 kJ. However, we need to adjust the stoichiometry of the reaction to match the formation of CuO2. Since we need 4 moles of CuO, we need to multiply the enthalpy change by the stoichiometric coefficient:
4 * (-288 kJ) = -1,152 kJ
The second reaction has a standard enthalpy change of 11 kJ. Here, we also need to adjust the stoichiometry to match the formation of CuO2. We can see that 1 mole of CuO2 is formed from 1 mole of Cu2O and 1/2 mole of O2. Therefore, we need to multiply the enthalpy change by the stoichiometric coefficient:
1 * (1/2 * 11 kJ) = 5.5 kJ
Now, we can sum up the enthalpy changes of the two reactions to calculate the overall standard enthalpy change for the formation of CuO2:
-1,152 kJ + 5.5 kJ = -1,146.5 kJ
Therefore, the standard enthalpy of formation of CuO2 is -1,146.5 kJ.
Standard enthalpy of formation is creating a compound from it's elements- you don't need the equation DrBob222.
The equation would read: Cu + O2 yields CuO2.
Then use Hess's law to figure it out.