Math Problem Statement

Phthalates used as plasticizers in rubber and plastic products are believed to act as hormone mimics in humans. The value of ΔHcomb for dimethylphthalate (C10H10O4) is –4685 kJ/mol. Assume 0.940  g of dimethylphthalate is combusted in a calorimeter whose heat capacity (Ccalorimeter) is 7.65  kJ/°C at 20.5 °C. What is the final temperature of the calorimeter?

Solution

To determine the final temperature of the calorimeter after the combustion of dimethylphthalate, we need to follow these steps:

1. Calculate the number of moles of dimethylphthalate (C10H10O4C_{10}H_{10}O_4) combusted.

Given:

  • Mass of dimethylphthalate, m=0.940gm = 0.940 \, \text{g}
  • Molar mass of dimethylphthalate, M=194.19g/molM = 194.19 \, \text{g/mol} (calculated using the molecular formula)

Moles of dimethylphthalate=mM=0.940g194.19g/mol0.00484mol\text{Moles of dimethylphthalate} = \frac{m}{M} = \frac{0.940 \, \text{g}}{194.19 \, \text{g/mol}} \approx 0.00484 \, \text{mol}

2. Calculate the heat released by the combustion of dimethylphthalate.

Given:

  • Heat of combustion, ΔHcomb=4685kJ/mol\Delta H_{\text{comb}} = -4685 \, \text{kJ/mol}

Heat released=moles×ΔHcomb=0.00484mol×(4685kJ/mol)22.68kJ\text{Heat released} = \text{moles} \times \Delta H_{\text{comb}} = 0.00484 \, \text{mol} \times (-4685 \, \text{kJ/mol}) \approx -22.68 \, \text{kJ}

3. Calculate the temperature change of the calorimeter.

Given:

  • Heat capacity of the calorimeter, Ccalorimeter=7.65kJ/°CC_{\text{calorimeter}} = 7.65 \, \text{kJ/°C}

The heat released by the reaction is absorbed by the calorimeter, causing an increase in its temperature.

q=Ccalorimeter×ΔTq = C_{\text{calorimeter}} \times \Delta T

ΔT=qCcalorimeter=22.68kJ7.65kJ/°C2.96°C\Delta T = \frac{q}{C_{\text{calorimeter}}} = \frac{-22.68 \, \text{kJ}}{7.65 \, \text{kJ/°C}} \approx -2.96 \, \text{°C}

4. Determine the final temperature of the calorimeter.

Given:

  • Initial temperature of the calorimeter, Tinitial=20.5°CT_{\text{initial}} = 20.5 \, \text{°C}

Since qq is negative, it indicates heat is released by the system, causing an increase in temperature.

Tfinal=Tinitial+ΔT=20.5°C+2.96°C23.46°CT_{\text{final}} = T_{\text{initial}} + \Delta T = 20.5 \, \text{°C} + 2.96 \, \text{°C} \approx 23.46 \, \text{°C}

So, the final temperature of the calorimeter is approximately 23.46°C.

Would you like more details or have any questions?

Here are five related questions to explore further:

  1. What would be the final temperature if twice the amount of dimethylphthalate was combusted?
  2. How would the final temperature change if the calorimeter had a heat capacity of 10 kJ/°C?
  3. What is the significance of the sign of ΔHcomb\Delta H_{\text{comb}} in this context?
  4. How do you calculate the molar mass of a compound like dimethylphthalate?
  5. What is the role of a calorimeter in measuring heat changes during chemical reactions?

Tip: When performing calorimetry calculations, always ensure that the heat capacity and temperature change are consistent in terms of units, to avoid errors in determining the final temperature.

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Math Problem Analysis

Mathematical Concepts

Calorimetry
Heat capacity
Thermochemistry

Formulas

q = C \times \Delta T
\Delta T = \frac{q}{C}

Theorems

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Suitable Grade Level

Advanced High School or College