A sample consisting of 0.025 mol of a solid compound is placed inside a metal cylinder. The cylinder was immersed in a water bath consisting of 10.00 kg of water at an original temperature of 25.00 °C. When the compound decomposes, the temperature of the water rises to 29.52 °C. Assume that the heat capacity of the cylinder is negligible. The specific heat of water is 4.184 J/(°C.g).
a. Calculate the heat evolved (in kJ) during the decomposition reaction, and from that determine the molar enthalpy change of the decomposition reaction.
b. The decomposition reaction may be written in this form:
C12O36H20N12 (s) → 12CO2 (g) + O2 (g) + 6N2 (g) + 10H2O (l)
In the enthalpy diagram below, the arrows represent standard enthalpy reactions. (3) denotes the standard enthalpy of decomposition of C12O36H20N12 (s). Provide an appropriate label for the arrows (1), (2) and (4).
(3) C12O36H20N12 (s) --> 12CO2 (g) + O2 (g) + 6N2 (g) + 10H2O(l)
(1) 12C (s) + 18O2 (g) + 6N2 (g) + 10H2 (g) --> 12CO2 (g) + 6O2 (g) + 6N2 (g) + 10H2 (g)
(2) 12CO2 (g) + 6O2 (g) + 6N2 (g) + 10H2 (g) --> 12CO2 (g) + O2 (g) + 6N2 (g) + 10H2O(l)
(4) 12C (s) + 18O2 (g) + 6N2 (g) + 10H2 (g) --> C12O36H20N12 (s)
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.)
q = mass water x specific heat water x (Tfinal-Tinitial).
Then q/0.025 mol = enthalpy/mol.
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waz up
To answer part (a) of the question, we need to calculate the heat evolved during the decomposition reaction and then use that value to determine the molar enthalpy change.
1. Calculate the heat evolved during the decomposition reaction:
We can use the equation q = mcΔT, where q represents the heat, m is the mass of water, c is the specific heat of water, and ΔT is the change in temperature.
Given:
Mass of water (m) = 10.00 kg = 10,000 g
Specific heat of water (c) = 4.184 J/(°C.g)
Change in temperature (ΔT) = 29.52 °C - 25.00 °C = 4.52 °C
Using the equation, q = (10,000 g) * (4.184 J/(°C.g)) * (4.52 °C), we can calculate the heat evolved.
2. Convert the heat evolved to kJ:
Since the specific heat of water is given in J/(°C.g) and the mass of water is given in grams, the heat is initially in Joules. To convert it to kilojoules, we divide by 1000.
Therefore, the heat evolved during the decomposition reaction is (10,000 g) * (4.184 J/(°C.g)) * (4.52 °C) / 1000 = 189.6288 kJ.
3. Determine the molar enthalpy change:
Given that there are 0.025 mol of the compound, we can divide the heat evolved by the number of moles to obtain the molar enthalpy change.
Molar enthalpy change = 189.6288 kJ / 0.025 mol = 7585.152 kJ/mol.
For part (b) of the question, we need to label the arrows (1), (2), and (4) on the enthalpy diagram.
Looking at the equation for the decomposition reaction:
C12O36H20N12 (s) → 12CO2 (g) + O2 (g) + 6N2 (g) + 10H2O (l)
- The arrow labeled (1) represents the formation of the reactants, 12C (s) + 18O2 (g) + 6N2 (g) + 10H2 (g), which are required to form the products.
- The arrow labeled (2) represents the combustion of the reactants, 12CO2 (g) + 6O2 (g) + 6N2 (g) + 10H2 (g), resulting in the formation of carbon dioxide, nitrogen gas, and water.
- The arrow labeled (4) represents the reverse reaction, where C12O36H20N12 (s) is formed from its constituent elements, 12C (s) + 18O2 (g) + 6N2 (g) + 10H2 (g).
For part (c) of the question, we need to use Hess's Law and the enthalpy diagram to determine the standard enthalpy of formation for C12O36H20N12 (s).
Hess's Law states that the overall enthalpy change for a reaction is equal to the sum of the enthalpy changes of the individual steps of the reaction.
Here's how we can use Hess's Law to determine the standard enthalpy of formation for C12O36H20N12 (s):
1. Start with the decomposition reaction:
C12O36H20N12 (s) → 12CO2 (g) + O2 (g) + 6N2 (g) + 10H2O (l)
2. Identify the steps in the enthalpy diagram that involve the formation of C12O36H20N12 (s):
- Step (4): 12C (s) + 18O2 (g) + 6N2 (g) + 10H2 (g) → C12O36H20N12 (s)
3. Determine the enthalpy change for step (4) by subtracting the enthalpy changes of the other two steps from the overall enthalpy change (3):
Enthalpy change for step (4) = Enthalpy change of step (3) - (Enthalpy change of step (1) + Enthalpy change of step (2))
4. Use the given values for the enthalpy changes in step (1), step (2), and the overall enthalpy change (3) to solve for the enthalpy change of step (4).
Once you have determined the enthalpy change for step (4), you can use the stoichiometry of the balanced equation to calculate the standard enthalpy of formation for C12O36H20N12 (s).