Hydrogen peroxide, H2O2, is a colorless liquid whose solutions are used as a bleach and an antiseptic. H2O2 can be prepared in a process whose overall change is the following.
H2(g) + O2(g) H2O2(l)
Calculate the enthalpy change using the following data.
H2O2(l) H2O(l) + 1/2 O2(g) ÄH = -98.0 kJ
2 H2(g) + O2(g) 2 H2O(l) ÄH = -571.6 kJ
Step reactions Given:
H2O2(l) –> H2O(l) + 1/2 O2(g) ∆H = -98.0 kJ
2 H2(g) + O2(g) –> 2 H2O(l) ∆H = -571.6 kJ
Reverse the first reaction and change the sign of the ∆H. Divide all the coefficients of the 2nd reaction and the ∆H by 2:
H2O(l) + 1/2 O2(g) –> H2O2(l) ∆H = +98.0 kJ
H2(g) + 1/2 O2(g) –> H2O(l) ∆H = -285.8 kJ
If you add the last two reactions you get:
H2(g) + O2(g) –> H2O2(l) (∆Hrxn = Sum of the two ∆H’s)
Well, I'm not really good with calculations, but let me give this a shot in my own humorous way.
So, in the first reaction, it's like H2 and O2 are getting together to form H2O2. It's like a match made in chemical heaven! And when they come together, they release some enthalpy, which is like the energy of their love.
And then in the second reaction, it's like H2 and O2 are breaking up, but luckily they find someone else - H2O - to be with. And when they do that, they release even more enthalpy, because they're so happy with their new partner.
Now, to calculate the overall enthalpy change, we need to do some math. We can just add up the enthalpy changes of the individual reactions to get the overall change.
So, first reaction: -98.0 kJ
Second reaction: -571.6 kJ
Overall change: -98.0 kJ + (-571.6 kJ) = -669.6 kJ
So, the overall enthalpy change is -669.6 kJ. That's like a roller coaster of emotions for those molecules!
To calculate the enthalpy change for the reaction H2(g) + O2(g) -> H2O2(l), we can use Hess's Law.
Hess's Law states that the overall enthalpy change for a reaction is equal to the sum of the enthalpy changes of the individual steps of the reaction.
In this case, we have two equations that can be combined to give the desired reaction:
1. H2O2(l) -> H2O(l) + 1/2 O2(g) ΔH = -98.0 kJ
2. 2 H2(g) + O2(g) -> 2 H2O(l) ΔH = -571.6 kJ
To combine these equations and cancel out common species, we'll reverse equation 1 and multiply equation 2 by 2:
1. -1(H2O(l) + 1/2 O2(g) -> H2O2(l)) ΔH = 98.0 kJ
2. 2(2 H2O(l) -> 2 H2(g) + O2(g)) ΔH = -1143.2 kJ
Now, we can add the two equations together to get the desired reaction:
2 H2(g) + O2(g) -> H2O2(l) ΔH = 98.0 kJ + (-1143.2 kJ) = -1045.2 kJ
Therefore, the enthalpy change for the reaction H2(g) + O2(g) -> H2O2(l) is -1045.2 kJ.
To calculate the enthalpy change for the given reaction, we need to use the concept of Hess's Law, which states that the enthalpy change for a reaction is independent of the pathway between the reactants and products and depends only on the initial and final states.
Given the following equations:
H2O2(l) -> H2O(l) + 1/2 O2(g) ΔH = -98.0 kJ (Equation 1)
2 H2(g) + O2(g) -> 2 H2O(l) ΔH = -571.6 kJ (Equation 2)
H2(g) + O2(g) -> H2O2(l) ΔH = ?
Our goal is to find the enthalpy change for the reaction in question.
First, we reverse Equation 1 since we need the reactants to match those in Equation 3:
H2O(l) + 1/2 O2(g) -> H2O2(l) ΔH = +98.0 kJ (Equation 1')
Then, we multiply Equation 1' by 2 to match the coefficients in Equation 2:
2 H2O(l) + O2(g) -> 2 H2O2(l) ΔH = +2(98.0 kJ) = +196.0 kJ (Equation 1'')
Now, we can add Equation 1'' to Equation 2, keeping in mind to reverse Equation 1'' to match the reactants:
(2 H2(g) + O2(g) -> 2 H2O(l)) + (2 H2O(l) + O2(g) -> 2 H2O2(l)) = 2 H2(g) + O2(g) -> 4 H2O2(l) ΔH = -571.6 kJ + 196.0 kJ = -375.6 kJ (Equation 3)
Finally, to find the enthalpy change for the reaction in question, we divide Equation 3 by 2:
(2 H2(g) + O2(g) -> 4 H2O2(l)) / 2 = H2(g) + 1/2 O2(g) -> 2 H2O2(l) ΔH = -375.6 kJ / 2 = -187.8 kJ
Therefore, the enthalpy change for the process H2(g) + O2(g) -> H2O2(l) is -187.8 kJ.