these are my three thermochemical equations
2Cu(s)+S(s) give Cu_2S(s) delta H -79.5 KJ
S(s)+O_2 give SO_2(g) -297 KJ
Cu_2S(s)+2O_2(g) give 2CuO(s)+SO_2(g)
how i calculate standard enthalpy of formation in Kj per mol of CuO(s)?
please don't give the answer only guid me please
Add equation 1 to the reverse of equation 2 to equation 3. That will give you twice the equation you want and you can take half of it later.
Add kJ/mole for equation to the negative of equation 2 to the kJ for equation 3. All of that together gives twice what you are looking for so divide the total by 2 and divide the equation by 2.
in front of last equation i forgat to wright the delta H .it is -527.5 KJ
That doesn't change anything. When you reverse an equation change the sign of delta H, otherwise leave it.
To calculate the standard enthalpy of formation (ΔHf) in kJ/mol of CuO(s), you can use the given thermochemical equations and apply the Hess's Law. The standard enthalpy of formation is defined as the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states.
To calculate the standard enthalpy of formation of CuO(s), you need to consider the following steps:
Step 1: Balancing the Equations
First, you need to balance the equations so that the desired compound, CuO(s), is represented.
2Cu(s) + S(s) → Cu2S(s) ΔH = -79.5 kJ
S(s) + O2(g) → SO2(g) ΔH = -297 kJ
Cu2S(s)+ 2O2(g) → 2CuO(s) + SO2(g)
Step 2: Manipulating the Equations
Next, you will manipulate the equations to align them in a way that allows you to cancel out common species, ultimately leading to the formation of CuO(s).
Reverse equation 1 and multiply by 2 to get:
2Cu2S(s) → 4Cu(s) + 2S(s) ΔH = +159 kJ
Add equation 2 to equation 3:
2Cu2S(s) + 2S(s) + 2O2(g) → 4Cu(s) + 2S(s) + 2O2(g) + SO2(g) ΔH = +159 kJ - 297 kJ
Simplify the equation:
2Cu2S(s) + 2O2(g) → 4Cu(s) + SO2(g) ΔH = -138 kJ
Step 3: Applying Hess's Law
As per Hess's Law, the sum of the enthalpy changes of a series of reactions equals the enthalpy change of the overall reaction.
Since you want to find the standard enthalpy of formation of CuO(s), set up the equation as follows:
2Cu(s) + O2(g) → 2CuO(s) ΔH = ?
By cancelling out common species from equation 3 and the newly formed equation, you get:
2Cu2S(s) + 2O2(g) → 4Cu(s) + SO2(g) ΔH = -138 kJ
2Cu(s) + O2(g) → 2CuO(s) ΔH = ?
Now, you can see that if you multiply equation 2 by 2 and subtract it from equation 3, you will get:
2Cu2S(s) → 2CuO(s) + SO2(g) ΔH = -138 kJ - 2(ΔHf of CuO)
Since ΔHf of Cu(s) and SO2(g) is zero, you're left with:
2Cu2S(s) → 2CuO(s) ΔH = -138 kJ - 2(ΔHf of CuO)
Finally, isolate ΔHf of CuO by rearranging the equation:
ΔHf of CuO = (-138 kJ) / 2
Thus, you can calculate the standard enthalpy of formation of CuO(s) in kJ/mol using the given equation and the values obtained from the other reactions.