Hydrogen peroxide is used as a bleachin agent, a disinfectant and medicinally as a topical antibacterial agent. Hydrogen peroxide decomposes according to the reacton
2H^2O^2(l)-->2H^2O(l)+o^2(g)
A. Given the standard free energy of formation values shown below, calculate the free energy change for this reaction.
Gf(H^2O^2)=-120.4kJ/mol
Gf(H^2O)=-237.1kJ/mol
For Further Reading
Chemistry - DrBob222, Tuesday, May 15, 2007 at 6:09pm
delta G reaction = delta G products - delta G reactants.
so the equation is delta G=-237.1kJ/mol-(-120.4)kJ/mol
and the free energy change is -116.7 kJ/mol.
Can you check my work please?
Would it not be 2 times that for delta G H2O2 and 2 times that for delta G H2O since two mols are involved for each? Otherwise it looks ok.
So should i times -237.1 and -120.4 by 2 to get an answer of -233.4 or is 116.7 correct?
delta Grxn = (2*-237.1)-(2*-120.4) = -233.4 kJ/rxn = -116.7 kJ/mol.
Sometimes it is tough to know which is the correct answer. In this case, however, the question asks for delta G for the REACTION; therefore, I think -233.4 is the correct answer because the reaction involves two mols. As a point of interest, however, I have seem many problems in texts that are not very consistent and they will give the reaction (for two mols such as this), then write to the side of the reaction, -116.7 kJ/mol and the student is supposed to pick up on that and multiply the stated value by 2 to obtain the reaction value. I hope this helps. Check my thinking.
Your thinking is correct. Since the reaction involves 2 moles of H2O2 and 2 moles of H2O, you need to multiply the standard free energy of formation values by 2 to get the correct answer.
So, the calculation would be:
ΔGrxn = (2 * (-237.1 kJ/mol)) - (2 * (-120.4 kJ/mol))
= -474.2 kJ/mol + 240.8 kJ/mol
= -233.4 kJ/mol
Therefore, the correct value for the standard free energy change of the reaction is -233.4 kJ/mol.