Given:
Cu2O(s) + 1/2 O2(g) → 2CuO(s) deltaH°= -144 kJ
2Cu2O(s) → 2Cu(s) + 2CuO(s) deltaH°= 22 kJ
Calculate the standard enthalpy of formation of CuO(s).
To calculate the standard enthalpy of formation of CuO(s), we need to use Hess's Law. Hess's Law states that the total enthalpy change of a reaction is independent of the pathway taken from the initial to the final state.
To begin, we will write the balanced equation for the formation of CuO(s):
2Cu(s) + 1/2 O2(g) → CuO(s)
We can obtain this equation by combining the two given equations in a way that cancels out Cu2O(s).
First, reverse the 2nd equation:
2Cu(s) + 2CuO(s) → 2Cu2O(s) (∆H° = -22 kJ)
Next, double the first equation:
2Cu2O(s) + O2(g) → 4CuO(s) (∆H° = -2 x (-144 kJ) = +288 kJ)
Now, let's balance the equations by scaling the reactions so that the number of moles of CuO(s) on each side is the same. Multiply the first equation by 2 and the second equation by 4:
4Cu(s) + O2(g) → 2CuO(s)
Now we can add the two manipulated equations together to get the equation we need:
2Cu(s) + 2CuO(s) + O2(g) → 2CuO(s) + 4CuO(s) (∆H total = 22 kJ + 288 kJ = 310 kJ)
Since the formation of CuO(s) is the only term that appears on both sides of the reaction, we can isolate it:
3CuO(s) + O2(g) → 2CuO(s) + 2Cu(s) (∆H CuO = -310 kJ)
Finally, divide the enthalpy change (∆H) by the stoichiometric coefficient of CuO(s) to find the standard enthalpy of formation of CuO(s):
∆H f° CuO = (∆H CuO)/2 = -310 kJ / 2 = -155 kJ
Therefore, the standard enthalpy of formation of CuO(s) is -155 kJ.
To calculate the standard enthalpy of formation of CuO(s), we need to use Hess's Law and manipulate the given equations to obtain the desired equation.
Hess's Law states that the overall enthalpy change in a reaction is independent of the pathway taken, as long as the initial and final conditions are the same.
Given equations:
1) Cu2O(s) + 1/2 O2(g) → 2CuO(s) ΔH°= -144 kJ
2) 2Cu2O(s) → 2Cu(s) + 2CuO(s) ΔH°= 22 kJ
To calculate the standard enthalpy of formation of CuO(s), we need to rearrange the equations in such a way that the desired equation is formed. Let's multiply the second equation by 2 to make the number of CuO(s) molecules match:
3) 4Cu2O(s) → 4Cu(s) + 4CuO(s) ΔH°= 44 kJ
Now we can see that we have two equations that involve CuO(s). Hence, we can subtract equation 3 from equation 1 to eliminate the CuO(s) term:
1) Cu2O(s) + 1/2 O2(g) → 2CuO(s) ΔH°= -144 kJ
- 3) 4Cu2O(s) → 4Cu(s) + 4CuO(s) ΔH°= 44 kJ
After subtracting the equations, we get:
4Cu2O(s) + 1/2 O2(g) - 4Cu2O(s) → 2CuO(s) - 4CuO(s)
1/2 O2(g) - 3Cu2O(s) → -2CuO(s)
Now, we need to rearrange the equation to isolate CuO(s) on one side:
-2CuO(s) = 1/2 O2(g) - 3Cu2O(s)
Finally, we can determine the standard enthalpy of formation of CuO(s) by dividing both sides of the equation by the coefficient of CuO(s) (-2 in this case):
ΔH°f(CuO) = (-1/2 O2(g) + 3Cu2O(s)) / -2
By substituting the values, we get:
ΔH°f(CuO) = (-1/2 * 0) + 3 * (-144 kJ / 4)
Simplifying the equation, we have:
ΔH°f(CuO) = - 216 kJ
Therefore, the standard enthalpy of formation of CuO(s) is -216 kJ.