Hydrogen peroxide can be prepared in several ways. One method is the reaction between hydrogen and oxygen, another method is the reaction between water and oxygen. Calculate the ΔG°rxn of each reaction below?

1) H2(g) + O2(g) <==> H2O2(l) ΔG°rxn=? (in kJ)
2) H2O(l) + 1/2O2(g) <==> H2O2(l) ΔG°rxn=? (in kJ)

Which method requires less energy under standard conditions?
Reaction 1
Reaction 2

I know reaction 1 is -120.4 kj and reaction 2 is 116.7 kj, but I'm not sure about the second part, will the second one require less energy because it has a positive energy?

Its reaction 1.

Well, let me clown around with the second part and shed some light on it! You are correct that reaction 1 has a ΔG°rxn of -120.4 kJ, while reaction 2 has a ΔG°rxn of 116.7 kJ. Now, pay attention, because this is where the funny business happens!

In both reactions, we are aiming to form hydrogen peroxide (H2O2). In reaction 1, we directly combine hydrogen (H2) and oxygen (O2) to form H2O2. Now, reaction 2 takes a comical twist! We begin with water (H2O) and only half of an oxygen molecule (1/2O2) to form H2O2. It's like making hydrogen peroxide with half-hearted oxygen!

But when it comes to the energy required, we need to look at the signs of the ΔG°rxn values. In reaction 1, the value is negative, indicating that it releases energy. In reaction 2, the value is positive, which means you'll need to put in some energy.

So, drumroll, please! The method that requires less energy under standard conditions is reaction 1! It's the winner of the "Least Energy Required" award. It's a bummer that reaction 2 needs a positive energy input, like giving a motivational speech to the oxygen molecule!

I hope that clears things up with a sprinkle of humor! Just remember, never underestimate the power of H2O2, the superhero of cleaning and disinfecting!

To determine which method requires less energy under standard conditions, you need to compare the magnitudes of the values for ΔG°rxn for each reaction. In this case:

1) H2(g) + O2(g) ⇌ H2O2(l) ΔG°rxn = -120.4 kJ
2) H2O(l) + 1/2O2(g) ⇌ H2O2(l) ΔG°rxn = 116.7 kJ

The magnitude of the value for ΔG°rxn represents the amount of energy released or consumed during the reaction. In this case, the magnitude of the value for reaction 1 is 120.4 kJ, while the magnitude of the value for reaction 2 is 116.7 kJ.

Therefore, reaction 2 requires less energy under standard conditions because the magnitude of its value for ΔG°rxn is smaller than that of reaction 1.

To calculate the ΔG°rxn (standard Gibbs free energy change) for each reaction, you can use the equation ΔG°rxn = ΣΔG°f(products) - ΣΔG°f(reactants), where ΔG°f is the standard Gibbs free energy of formation.

Let's calculate the ΔG°rxn for each reaction:

1) H2(g) + O2(g) ⇌ H2O2(l)
ΔG°f(H2O2) = -120.4 kJ/mol (given)
ΔG°rxn = ΣΔG°f(products) - ΣΔG°f(reactants)
= ΔG°f(H2O2) - (ΔG°f(H2) + ΔG°f(O2))
= -120.4 - (0 + 0)
= -120.4 kJ/mol

2) H2O(l) + 1/2O2(g) ⇌ H2O2(l)
ΔG°f(H2O2) = -120.4 kJ/mol (given)
ΔG°f(H2O) = -237.1 kJ/mol (known value)
ΔG°rxn = ΣΔG°f(products) - ΣΔG°f(reactants)
= ΔG°f(H2O2) - (ΔG°f(H2O) + 1/2ΔG°f(O2))
= -120.4 - (-237.1 + 0)
= 116.7 kJ/mol

Therefore, the values you have provided are correct.

Now, let's compare the ΔG°rxn for each reaction. The higher the absolute value of ΔG°rxn, the more energy is required for the reaction to occur under standard conditions.

Reaction 1 has a ΔG°rxn of -120.4 kJ/mol, indicating that the reaction is exergonic (releasing energy).

Reaction 2 has a ΔG°rxn of 116.7 kJ/mol, indicating that the reaction is endergonic (requiring energy).

Based on these values, we can conclude that Reaction 1 requires less energy under standard conditions because its ΔG°rxn is smaller in absolute value.

Reaction 1 requires more energy because when delta G is a negative number you know your reaction is spontaneous.Just like when delta G is positive we know the reaction is non-spontaneous. If a reaction takes place spontaneously, it makes more sense for it to use less energy then a reaction that is non-spontaneous.