I don't know why it was marked wrong. This is what it said: Where is the 5.64 coming from? (thank you)
HINT
To balance the equation, make sure that the number of each atom on the reactant side is the same as the number on the product side. Should the enthalpy change for this reaction have a positive sign or a negative sign? From the balanced thermochemical equations, how many moles of C12H22O11 (sucrose) are thermochemically equivalent to 5.64 x 10^3 kJ? Use this equivalent quantity as a conversion factor to calculate how much heat is released per gram of sucrose oxidized . Make sure all your units cancel out and the signs are correct in your calculations.
Sucrose (C12H22O11, table sugar) is oxidized in the body by O2 via a complex set of reactions that ultimately produces CO2 (g) and H2O (g) and releases 5.16x10^3 kJ/mol sucrose.
a) Write a balanced thermochemical equation for this reaction. Make sure to include the physical state of each reactant and product........>>>>>>..........
b)Enter the answer for Delta Hrxn ........................answer in SCIENTIFIFIC NOTATION in kJ Moles
c) How much heat is released per gram of sucrose oxidized? ...................... answer in kJ/g
is b.....-5.16 x 10^3
is .c..............15.1 kJ/g ??? thank you.•chemistry - DrBob222, Sunday, October 18, 2015 at 1:32pm
C12H22O11(s) + 12O2(g) ==> 12CO2(g) + 11H2O(g) dH = -5.16E-3 kJ
b is -5.16E+3 kJ/mol
c is right
Dr bob...you were right again! Thank you, it was a typo😊
To understand where the value of 5.64 is coming from, we need to follow the steps provided in the question.
Step 1: Write the balanced thermochemical equation for the reaction.
The given equation is: C12H22O11(s) + 12O2(g) → 12CO2(g) + 11H2O(g)
Step 2: Determine the enthalpy change (ΔHrxn) for the reaction.
From the given information, we know that the enthalpy change for the reaction is -5.16 x 10^3 kJ/mol. Therefore, the answer for part b is -5.16 x 10^3 kJ/mol.
Step 3: Calculate the thermochemical equivalent of sucrose.
To determine how many moles of C12H22O11 (sucrose) are thermochemically equivalent to 5.64 x 10^3 kJ, we can use the conversion factor provided in the equation. The equation tells us that 5.16 x 10^3 kJ is equivalent to one mole of sucrose. Therefore, we can set up a proportion:
(5.16 x 10^3 kJ) / (1 mol) = (5.64 x 10^3 kJ) / (x mol)
Cross-multiplying and solving for x, we find:
x = (5.64 x 10^3 kJ) / (5.16 x 10^3 kJ/mol) ≈ 1.09 mol
So, the thermochemical equivalent of 5.64 x 10^3 kJ is approximately 1.09 moles of C12H22O11.
Step 4: Calculate the heat released per gram of sucrose oxidized.
To find the heat released per gram of sucrose, we need to convert the moles of sucrose to grams. The molar mass of sucrose (C12H22O11) can be calculated as follows:
(12 x atomic mass of C) + (22 x atomic mass of H) + (11 x atomic mass of O)
= (12 x 12.01 g/mol) + (22 x 1.008 g/mol) + (11 x 16.00 g/mol)
= 342.34 g/mol
Now, we can set up another proportion to calculate the heat released per gram:
(1.09 mol) / (342.34 g) = (x mol) / (1 g)
Cross-multiplying and solving for x, we find:
x ≈ 0.00319 mol/g
Finally, to convert moles per gram to kilojoules per gram, we multiply by the heat of reaction (ΔHrxn) from part b:
Heat released per gram = (0.00319 mol/g) * (-5.16 x 10^3 kJ/mol)
≈ -15.1 kJ/g
Therefore, the answer for part c is approximately -15.1 kJ/g.