The human body “burns” glucose (C6H12O6) for energy.
a. Write a balanced chemical reaction for the complete combustion of glucose to CO2(g)
and H2O(l) such that the stoichiometric coefficient for glucose is 1.
b. When 1.00 g of glucose are completely burned at 298.15 K and 1 bar of pressure,
15.7 kJ are released. Calculate the ÄH0
f
of glucose.
C6H12O6 + 6O2 ==> 6CO2 + 6H2O + 15.7 kJ
dHfrxn = (n*dHf products) - (n*dHf reactants)
15.7 kJ/mol = [(6*dH CO2)+(6*dH H2O)] - (0 + dH glucose).
Solve for dH glucose.
Excellent
a. The balanced chemical reaction for the complete combustion of glucose (C6H12O6) to CO2(g) and H2O(l) is:
C6H12O6 + 6 O2 -> 6 CO2 + 6 H2O
b. To calculate the standard enthalpy change (ΔH°f) of glucose, we need to use the following equation:
ΔH°f = q / n
Where ΔH°f is the standard enthalpy change, q is the heat energy released, and n is the number of moles.
First, we need to calculate the number of moles of glucose:
Molar mass of glucose (C6H12O6) = 180.16 g/mol
Number of moles (n) = mass / molar mass
n = 1.00 g / 180.16 g/mol ≈ 0.00555 mol
Next, we can plug in the values into the equation:
ΔH°f = 15.7 kJ / 0.00555 mol ≈ -2824.82 kJ/mol
Therefore, the standard enthalpy change (ΔH°f) of glucose is approximately -2824.82 kJ/mol.
a. To write a balanced chemical reaction for the complete combustion of glucose to CO2(g) and H2O(l), we need to determine the reactant and product molecules. Glucose (C6H12O6) is reacting with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). The balanced equation will have equal numbers of atoms of each element on both sides.
Glucose (C6H12O6) + Oxygen (O2) → Carbon Dioxide (CO2) + Water (H2O)
To balance the equation, we need to determine the stoichiometric coefficients for each molecule by ensuring that the number of atoms of each element is equal on both sides of the equation.
C6H12O6 + 6O2 → 6CO2 + 6H2O
b. To calculate the ΔH°f of glucose, we need to use the equation:
ΔH°f of glucose = ΔH°f of CO2 + ΔH°f of H2O - ΔH°f of reactants
To calculate the ΔH°f of CO2 and H2O, we can refer to the standard enthalpy of formation (ΔH°f) values for these compounds. The enthalpy of formation of glucose is not given, so we'll need to calculate it using the given information.
Given:
Mass of glucose (m) = 1.00 g
Energy released (ΔH) = 15.7 kJ
To calculate the ΔH°f of glucose, we'll use the equation:
ΔH°f = ΔH / moles of glucose
First, we need to determine the number of moles of glucose.
Molar mass of glucose (C6H12O6) = 6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol) = 180.18 g/mol
Moles of glucose = mass of glucose / molar mass = 1.00 g / 180.18 g/mol ≈ 0.00555 mol
Now we can calculate the ΔH°f of glucose:
ΔH°f of glucose = 15.7 kJ / 0.00555 mol ≈ 2829 kJ/mol
Therefore, the ΔH°f of glucose is approximately 2829 kJ/mol.