Consider the following three reactions:
Ca(s) + 2H2O(l) ---> Ca2+(aq) + 2OH^- (aq) + H2(g)
CaO(s) + H2O(l) ---> Ca2+ (aq) + 2OH^- (aq)
H2(g) + 1/2O2 (g) ----> H2O (l)
Delta H = 430.75 kJ
Delta H = 81.55 kJ
Delta H = 285.63 kJ
Determine Delta H for the overall reaction:
Ca(s) + 1/2O2 (g) ---> CaO(s)
Add eqn 1 to eqn 3 to the reverse of eqn 2. Add the dH values; when you reverse the rxn change the sign of dH for that rxn.
Post your work if you get stuck.
To determine the ΔH for the overall reaction, we can use the Hess's Law, which states that the ΔH of a reaction is the sum of the ΔH values of the individual reactions that make up the overall reaction.
First, we need to balance the equations and reverse the second equation to use it in the calculation:
2Ca(s) + 2H2O(l) ---> 2Ca2+(aq) + 4OH^-(aq) + H2(g) (multiply the first equation by 2)
CaO(s) + H2O(l) ---> Ca2+(aq) + 2OH^-(aq) (reverse the second equation)
Now let's combine the equations:
2Ca(s) + 2H2O(l) + CaO(s) ---> 2CaO(s) + 2Ca2+(aq) + 6OH^-(aq) + H2(g)
By canceling out the Ca2+ and OH^- ions on both sides, we have:
2Ca(s) + 2H2O(l) + CaO(s) ---> 2CaO(s) + 3H2(g)
Now, let's add up the ΔH values for the individual reactions:
ΔH1 = -430.75 kJ (first reaction)
ΔH2 = 81.55 kJ (reversed second reaction)
ΔH3 = 285.63 kJ (third reaction)
Finally, to calculate the overall ΔH, we subtract the sum of ΔH2 and ΔH3 from ΔH1:
ΔHoverall = ΔH1 - ΔH2 - ΔH3
ΔHoverall = -430.75 kJ - 81.55 kJ - 285.63 kJ
ΔHoverall = -798.93 kJ
Therefore, the ΔH for the overall reaction: Ca(s) + 1/2O2(g) ---> CaO(s) is -798.93 kJ.
To determine ΔH for the overall reaction: Ca(s) + 1/2O2 (g) ---> CaO(s), you need to use the concept of Hess's Law.
Hess's Law states that the enthalpy change of a reaction can be calculated by summing up the enthalpy changes of the individual reactions that lead to the overall reaction.
In this case, you have three reactions with their respective enthalpy changes (ΔH values) provided:
Reaction 1: Ca(s) + 2H2O(l) ---> Ca2+(aq) + 2OH^-(aq) + H2(g) (ΔH = 430.75 kJ)
Reaction 2: CaO(s) + H2O(l) ---> Ca2+(aq) + 2OH^-(aq) (ΔH = 81.55 kJ)
Reaction 3: H2(g) + 1/2O2(g) ----> H2O(l) (ΔH = 285.63 kJ)
To calculate the ΔH for the desired overall reaction, you need to manipulate and combine the given reactions in a way that cancels out the common species and leaves you with the desired reaction.
First, consider reversing Reaction 1 and multiplying it by 2:
2(Ca2+(aq) + 2OH^-(aq) + H2(g)) ---> 2Ca(s) + 4H2O(l) (ΔH = -2 * 430.75 kJ = -861.50 kJ)
Next, combine this with Reaction 2, ensuring that species Ca2+ and OH^- are canceled out:
2CaO(s) + 2H2O(l) ---> 2Ca2+(aq) + 4OH^-(aq) (ΔH = 2 * 81.55 kJ = 163.10 kJ)
Now, you have:
2CaO(s) + 2H2O(l) - 2(Ca2+(aq) + 2OH^-(aq) + H2(g)) ---> 2Ca(s) + 4H2O(l)
Simplifying:
2CaO(s) + H2O(l) - Ca(s) ---> 2Ca(s) + 3H2O(l) (ΔH = 163.10 kJ - 861.50 kJ = -698.40 kJ)
Finally, consider combining this with Reaction 3, making sure H2O(l) is canceled out:
2CaO(s) + H2(g) - 1/2O2(g) - Ca(s) ---> 2Ca(s)
This gives you:
Ca(s) + 1/2O2(g) ---> CaO(s) (ΔH = -698.40 kJ + 285.63 kJ = -412.77 kJ)
Therefore, the ΔH for the overall reaction: Ca(s) + 1/2O2(g) ---> CaO(s) is -412.77 kJ.