c. Use the enthalpy diagram provided above and apply Hess’s Law to determine the standard enthalpy of formation for C12O36H20N12 (s) using the results from part (a) and the following values:
The standard enthalpy of formation of gaseous carbon dioxide is -393.5 kJ/mol
The standard enthalpy of formation of liquid water is -286 kJ/mol
(Strictly speaking, standard enthalpy is defined for a reaction that occurs at 25 °C, but for the purpose of this question you may ignore the difference in temperature.)
Basically, this is as follows:
delta Hrxn = (delta H products) - (delta H reactants)
Where delta H products and delta H reactants follow this set up.
#moles*delta H product 1 + # moles*delta H product 2.....
To determine the standard enthalpy of formation (ΔHf) for C12O36H20N12 (s), we can use the concept of Hess's Law.
Hess's Law states that the change in enthalpy for a reaction is constant, regardless of the number of steps or intermediates involved, as long as the initial and final states remain the same.
Here's how we can apply Hess's Law:
1. Write the balanced chemical equation for the formation of C12O36H20N12 (s).
C12H22O11 (s) + 36 CO2 (g) + 12 H2O (l) + 6 N2 (g) → C12O36H20N12 (s)
2. Break down the formation equation into a set of simpler reactions for which the standard enthalpies of formation are given:
(a) C12H22O11 (s) + 12 O2 (g) → 12 CO2 (g) + 11 H2O (g)
(b) 12 CO2 (g) + 12 H2O (g) → C12O36H20N12 (s)
(c) 12 O2 (g) + 6 N2 (g) → 6 N2O (g)
3. Apply Hess's Law by manipulating the given reactions to obtain the formation reaction:
(a) C12H22O11 (s) + 12 O2 (g) → 12 CO2 (g) + 11 H2O (g) (reverse the reaction to get the required product on the right-hand side)
ΔH1 = -ΔHf(C12H22O11)
(b) 12 CO2 (g) + 12 H2O (g) → C12O36H20N12 (s) (reverse the reaction to get the required product on the right-hand side)
ΔH2 = -ΔHf(C12O36H20N12)
(c) 12 O2 (g) + 6 N2 (g) → 6 N2O (g) (keep the reaction as it is)
ΔH3 = ΔHf(N2O) x 6
4. Add the reactions together to cancel out common compounds:
ΔH1 + ΔH2 + ΔH3 = 0
-ΔHf(C12H22O11) + (-ΔHf(C12O36H20N12)) + (ΔHf(N2O) x 6) = 0
5. Rearrange the equation to solve for ΔHf(C12O36H20N12):
-ΔHf(C12O36H20N12) = ΔHf(C12H22O11) + (ΔHf(N2O) x 6)
6. Substitute the given standard enthalpy values:
-ΔHf(C12O36H20N12) = ΔHf(C12H22O11) + (-393.5 kJ/mol x 6)
-ΔHf(C12O36H20N12) = ΔHf(C12H22O11) - 2361 kJ/mol
7. Calculate the standard enthalpy of formation of C12O36H20N12 (s):
ΔHf(C12O36H20N12) = -(-ΔHf(C12O36H20N12))
Remember to use the given standard enthalpy values for C12H22O11 and N2O to complete the calculation and obtain the final answer.
To determine the standard enthalpy of formation for C12O36H20N12 (s), we can use Hess's Law, which states that the enthalpy change of a reaction is independent of the path taken in reaching the final state, as long as the initial and final conditions are the same.
Hess's Law allows us to calculate the overall enthalpy change of a reaction by combining two or more reactions with known enthalpy changes. In this case, we will use the enthalpy changes from part (a), the enthalpies of formation of gaseous carbon dioxide and liquid water.
The chemical equation for the formation of C12O36H20N12 (s) can be represented as follows:
12CO2(g) + 18H2O(l) + 6N2(g) → C12O36H20N12(s)
Now, let's break down the given equation into smaller steps where we know the enthalpy changes:
Step 1: Formation of carbon dioxide (CO2)
C(graphite) + O2(g) → CO2(g) ΔH = -393.5 kJ/mol
Step 2: Formation of water (H2O)
H2(g) + 1/2O2(g) → H2O(l) ΔH = -286 kJ/mol
Step 3: Combining the equations and canceling out common species to obtain the desired equation
12C(graphite) + 36O2(g) + 20H2(g) + 6N2(g) → C12O36H20N12(s)
By adding the enthalpy changes of the individual steps, we can determine the overall enthalpy change:
ΔH = 12(-393.5) + 18(-286) + 6(0) - 0
Simplifying:
ΔH = -4722 kJ/mol
Therefore, the standard enthalpy of formation for C12O36H20N12(s) is -4722 kJ/mol.