Given H2(g) + (1/2)O2(g) --> H2O, ∆Hº = -286 kJ/mol, determine the standard enthalpy change for the reaction 2H2O(l) --> 2H2(g) + O2(g).
Just checking... should end up with ∆Hº = 572 kJ/mol right??
Yes and no. Yes, the correct number is 572 kJ/reaction (not /mol since it's for two mols) AND I don't think you should call it dHo. dHo is -285 kJ/mol is for the H2 + 1/2 O2 ==> H2O. When you turn it around it becomes for the reaction and multiplying by 2 makes is for two mols. I know I'm picky but I'm earning my money. :-). Since I'm a volunteer there is no money but you get the point.
Well, I always like to check whenever it comes to chemistry, just to make sure we don't end up with an explosive situation! So, let's take a closer look at this reaction.
We know that the first reaction, H2(g) + (1/2)O2(g) --> H2O, has a standard enthalpy change (∆Hº) of -286 kJ/mol.
Now, the second reaction, 2H2O(l) --> 2H2(g) + O2(g), is the reverse of the first reaction. So, if we flip the first reaction, we also have to flip the sign of the enthalpy change.
Therefore, the standard enthalpy change for the second reaction, 2H2O(l) --> 2H2(g) + O2(g), would be +286 kJ/mol.
So, drum roll, please... ∆Hº = +286 kJ/mol! I hope this brings some light (and heat) to your question!
To determine the standard enthalpy change for the reaction 2H2O(l) → 2H2(g) + O2(g), you can use the given standard enthalpy change for the reaction H2(g) + (1/2)O2(g) → H2O, which is ∆Hº = -286 kJ/mol.
The reaction you are given is the reverse of the reaction H2(g) + (1/2)O2(g) → H2O. To obtain the standard enthalpy change for the reverse reaction, you need to multiply the standard enthalpy change by -1.
Therefore, the standard enthalpy change for the reverse reaction is:
-(-286 kJ/mol) = 286 kJ/mol
Since the reverse reaction is the same as the reaction you are asked to determine the standard enthalpy change for, you can conclude that the standard enthalpy change for the reaction 2H2O(l) → 2H2(g) + O2(g) is ∆Hº = 286 kJ/mol.
So, the correct answer is ∆Hº = 286 kJ/mol, not 572 kJ/mol.
To determine the standard enthalpy change (∆Hº) for the reaction 2H2O(l) → 2H2(g) + O2(g), you can use Hess's law. Hess's law states that if a reaction can be expressed as the sum of other reactions, then the enthalpy change of the original reaction is equal to the sum of the enthalpy changes of the other reactions.
Here, we can consider the given reaction, H2(g) + (1/2)O2(g) → H2O, with a standard enthalpy change of -286 kJ/mol.
First, let's reverse this reaction:
H2O → H2(g) + (1/2)O2(g)
Since the reaction is reversed, the sign of the enthalpy change also changes. Therefore, the enthalpy change for the reversed reaction is +286 kJ/mol.
Now, if we double the reversed reaction, we get:
2H2O → 2H2(g) + O2(g)
Doubling the reaction also doubles the enthalpy change. Therefore, the enthalpy change for the reaction 2H2O → 2H2(g) + O2(g) is 2 × 286 kJ/mol, which equals 572 kJ/mol.
So, you are correct. The standard enthalpy change for the reaction 2H2O(l) → 2H2(g) + O2(g) is indeed ∆Hº = 572 kJ/mol.