Use standard heats of formation in Appendix L to calculate standard enthalpy changes for the following reactions.
1.45 x 10-2 mol of carbon is oxidized to CO2(g)
Please help me!
Look up the heat of formation for CO2 in the table provided to you. My table has something like 393.5 kJ/mol. That reaction is for
C + O2 ==> CO2
delta H reaction is (n*deltaHf products) - (n*deltaHf reactants) =
(n*deltaHf CO2) - [(n*deltaHf C) + (n*deltaHf O2)]
delta Hrxn = -393.5 -(+0) = -393.5 kJ/mol.
You have 1.45E-2 mol instead of 1 mole; therefore, delta H rxn = -393.5 kJ/mol x 1.45E-2 mol = ?? kJ.
Thank you! but does n in (n*deltaHf products) - (n*deltaHf reactants) mean amount? What amount is it talkinging about?
To calculate the standard enthalpy change for the reaction of carbon being oxidized to CO2(g), we need to use the standard heats of formation from Appendix L. The standard heat of formation (∆Hf) is the change in enthalpy when one mole of a substance is formed from its constituent elements in their standard states at 25 degrees Celsius and one atmosphere pressure.
First, we need to identify the reactants and products in the reaction. In this case, the reactant is carbon (C), and the product is carbon dioxide gas (CO2(g)).
The balanced chemical equation for this reaction is:
C(s) + O2(g) -> CO2(g)
Next, we need to determine the standard heat of formation for each species involved in the reaction. According to Appendix L:
∆Hf(C(s)) = 0 kJ/mol (Standard heat of formation for carbon in its standard state)
∆Hf(CO2(g)) = -393.5 kJ/mol (Standard heat of formation for carbon dioxide gas)
Now, we can use these values to calculate the standard enthalpy change (∆H) for the reaction:
∆H = ∑ ∆Hf(products) - ∑ ∆Hf(reactants)
∆H = ∆Hf(CO2(g)) - ∆Hf(C(s))
∆H = -393.5 kJ/mol - 0 kJ/mol
∆H = -393.5 kJ/mol
Therefore, the standard enthalpy change for the reaction of carbon being oxidized to CO2 is -393.5 kJ/mol. This indicates that the reaction is exothermic, releasing energy in the form of heat.