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:
2C57H110O6 (s) + 163O2(g) --> 114CO2(g) + 110H2O(l)
and my work:
DHrxn = [(114x -393.5)+(110x -285.840)] - [(2x -390.7)+(163x0)] = -35120KJ/mol
It is exothermic
I am having problems with this question. I keep getting different answers. Is this one right?
The equation isn't balanced. You have (163*2)+(2*6) = 338 O on the left and (114*2) + 110*2) = 448 on the right. You can make it balance with 55 H2O and not 110 H2O.
If I use 55 for H2O, then the H isn't balanced
Yes it is.
2C57H110O6 (s) + 163O2(g) --> 114CO2(g) + 55H2O(l)
You have 110 H on the left and 110 on the right.
55x2 = 110 on the right
but have 220 on the left
I am really embarrassed. I completely ignored that coefficient of 2 for the first material. Ok so the equation is ok. Let me recalculate the dH.
For this one I actually think it's -75520 KJ/mol
Also, could you please check my question with 578 g of C57H110O6 on page two?
I don't get 35,120. I get 75,520 for the balanced equation. If you take the question literally you divide by 2 but that gives my number still higher. I normally don't divide by 2 UNLESS the problem states that they want kJ/mol but I don't know the context of your questions so you may be right to divide by 2. However, we still differ by about 2600.
I don't understand the dividing by 2. Could you explain more?
To calculate the ΔHrxn (enthalpy change of reaction) using the given ΔHf (standard enthalpy of formation) values, you need to follow the steps correctly. Let's break it down:
1. Start by balancing the chemical equation correctly. You've done this step correctly:
2C57H110O6 (s) + 163O2(g) → 114CO2(g) + 110H2O(l)
2. Next, determine the ΔHrxn by calculating the sum of the products minus the sum of the reactants, multiplied by their stoichiometric coefficients.
ΔHrxn = [(114 × -393.5 kJ/mol) + (110 × -285.840 kJ/mol)] - [(2 × -390.70 kJ/mol) + (163 × 0 kJ/mol)]
= [-44811 + (-31442.4)] - [-781.4 + 0]
= -76253.4 kJ/mol
Therefore, the correct ΔHrxn for the reaction is -76253.4 kJ/mol. Since it is negative, the reaction is exothermic, as you correctly stated.
Make sure to double-check your calculations to ensure the correct values were used and that no errors were made during the calculations.