consider the following changes at constant temperature & pressure
H2O(s) --yields-- H20(l); Delta H1
H2O(l) --yields-- H2O(g); Delta H2
H2O(g) --yields-- H2O(s); Delta H3
Using Hess's law, the sum of the deltas is
a. equal to 0
b. sometimes greater than zero, sometimes less than zero
c. less than 0
d. cannot be determined without the numerical values of delta H
e. greater than 0
I picked e because delta 1 looks to be an endothermic reaction, delta 2 also endothermic, and delta 3 is exothermic reaction. But the answer is equal to 0. Why?
dHsub = dHfusion + dHvaporization
So you add dHfusion to dHvap and you end up with dHsub which means that if you complete the cycle you end you with zero.
To determine the correct answer, you need to apply Hess's law. Hess's law states that the overall enthalpy change of a reaction is independent of the pathway taken. This means that the sum of the enthalpy changes for a series of reactions will equal the enthalpy change of the desired reaction.
In this case, you have three reactions:
1. H2O(s) → H2O(l); ΔH1
2. H2O(l) → H2O(g); ΔH2
3. H2O(g) → H2O(s); ΔH3
To find the overall enthalpy change, you need to cancel out the intermediate states. Notice how ΔH1 and ΔH3 have opposite signs and represent the opposite processes of converting between solid and gas via the liquid phase. To cancel them out, you need to reverse one of them. Let's reverse ΔH3:
3. H2O(s) ← H2O(g); -ΔH3
Now, you can sum up the three reactions:
1. H2O(s) → H2O(l); ΔH1
2. H2O(l) → H2O(g); ΔH2
3. H2O(s) ← H2O(g); -ΔH3
Adding these equations together, you get:
H2O(s) → H2O(l) → H2O(g) → H2O(s)
The intermediate's cancel out, and you are left with the overall reaction:
H2O(s) → H2O(s)
Since the reactant and product are the same, the enthalpy change for this overall reaction is 0 (zero).
Therefore, the correct answer is (a) equal to 0.
The answer is actually a. equal to 0.
Hess's law states that the overall enthalpy change of a reaction is independent of the pathway taken. In other words, if you can express a desired reaction as a combination of other reactions, you can add up the individual enthalpy changes to find the overall enthalpy change.
In this case, let's rearrange the equations:
H2O(l) --yields-- H2O(s)
H2O(g) --yields-- H2O(l)
H2O(s) --yields-- H2O(g)
By changing the direction of the reactions, we can see that the second and third reactions are the reverse of each other. If we sum up these three equations, we get:
H2O(s) --yields-- H2O(l); ΔH1
H2O(l) --yields-- H2O(g); ΔH2
H2O(s) --yields-- H2O(g); ΔH3(reversed)
We can cancel out the intermediate H2O(l) by adding the first and second equation:
H2O(s) --yields-- H2O(g); ΔH1 + ΔH2
Then, by adding the reverse of the third equation:
H2O(s) --yields-- H2O(g); ΔH1 + ΔH2 - ΔH3(reversed)
Since ΔH3 is the reverse of ΔH3(reversed), they cancel out and we are left with:
H2O(s) --yields-- H2O(g); ΔH1 + ΔH2 - ΔH3 = 0
Therefore, at constant temperature and pressure, the sum of the ΔH values is equal to 0, as per Hess's law.