Calculate the DHrxn of
C57H110O6 (s) + O2(g) --> CO2(g) + H2O(l)
with DHf.
C57H110O6(s) -390.70
O2(g) 0
CO2(g) -393.5
H2O(l) -285.840
I balanced the equation and my work:
DHrxn = [224mol(-393.5+(-285.840))] - [165mol(-390.70+0)] = -87706.66 KJ/mol
Is this right? If not, what did I do wrong?
Again, here, I don't think the equation is balanced.
To calculate the DHrxn of a reaction using the enthalpy of formation (DHf) values, you need to consider the stoichiometric coefficients of the reactants and products. Let's go through the calculation step by step for the given reaction:
1. First, balance the equation:
C57H110O6 (s) + O2(g) --> 57CO2(g) + 55H2O(l)
2. Write down the given DHf values:
C57H110O6(s): -390.70 kJ/mol
O2(g): 0 kJ/mol
CO2(g): -393.5 kJ/mol
H2O(l): -285.840 kJ/mol
3. Calculate the DHrxn using the DHf values and stoichiometric coefficients:
DHrxn = [57 mol(-393.5 kJ/mol) + 55 mol(-285.840 kJ/mol)] - [1 mol(-390.70 kJ/mol) + 57 mol(0 kJ/mol)]
DHrxn = [-22406.50 kJ + (-15721.20 kJ)] - [-390.70 kJ]
DHrxn = -38127.70 kJ - (-390.70 kJ)
DHrxn = -37737 kJ/mol
Your calculation result of -87706.66 kJ/mol is incorrect. The correct DHrxn for the given reaction is -37737 kJ/mol.
Please note that the given DHf values are assumed to be correct, and the values may vary depending on the source and reference used for the enthalpy of formation values.