Calculate the DHrxn of
C6H12O6 (s) + O2 (g) --> CO2 (g) + H2O (l)
with DHf.
C6H12O6(s) -1260.04
O2(g) 0
CO2(g) -393.5
H2O(l) -285.840
I balanced the equation:
C6H12O6 (s) + 6O2 (g) --> 6CO2 (g) + 6H2O (l)
and my work:
DHrxn = [(6x -393.5)+(6x -285.840)] - [(1x -1260.04)+(6x0)] = -2552 KJ/mol
It is exothermic
I ma having problems with this question. I keep getting different answers. Is this one right?
I think it is set up right but I don't get that answer. It is exothermic and the equation is balanced.
how would I enter it into my calculator? I've entered it multiple ways and gotten different answers for each
If you show each step perhaps I can figures out what you're doing wrong.
For example:
6*-393.5 = ?
6*-285.84 = ?
Total = ?
etc.
-2816 KJ/mol?
That's what I obtained.
Thank you. Sorry for all the questions.
To calculate the ΔHrxn (enthalpy change) for the given reaction, you need to use the given ΔHf (standard enthalpy of formation) values for each compound involved in the reaction.
First, let's double-check your balanced equation:
C6H12O6 (s) + 6O2 (g) --> 6CO2 (g) + 6H2O (l)
Now, let's calculate the ΔHrxn using the ΔHf values:
ΔHrxn = [(6 × ΔHf(CO2)) + (6 × ΔHf(H2O))] - [(ΔHf(C6H12O6)) + (6 × ΔHf(O2))]
Substituting the given ΔHf values:
ΔHrxn = [(6 × (-393.5 kJ/mol)) + (6 × (-285.840 kJ/mol))] - [(-1260.04 kJ/mol) + (6 × 0 kJ/mol)]
ΔHrxn = [(-2361 kJ/mol) + (-1715.04 kJ/mol)] - [(-1260.04 kJ/mol)]
ΔHrxn = -4076.04 kJ/mol + 1260.04 kJ/mol
ΔHrxn = -2816 kJ/mol
So the correct value for ΔHrxn is -2816 kJ/mol. The negative sign indicates that the reaction is exothermic.
Make sure to double-check your calculations and ensure that you are using the correct ΔHf values for each compound.