Math Problem Statement

A student performed a calorimetry experiment. The neutralization reaction that took place during the experiment is represented by the equation below. HNO3(aq) + LiOH(aq) → LiNO3(aq) + HOH(I Numeric Response 2. 61.1 mL of 1.00 mol/L HNO3(aq) was combined with 78.0 mL of 1.00 mol/L LiOH(aq). The initial temperature of both solutions was 22.90°C and the highest temperature recorded was 29.00°C. Using this data, the molar enthalpy of neutralization of the nitric acid is +/..

Solution

To determine the molar enthalpy of neutralization of nitric acid (ΔHneut\Delta H_{\text{neut}}), we can use the data from the calorimetry experiment. The following steps outline the process:

  1. Determine the heat absorbed or released by the solution (qq): The heat change qq can be calculated using the formula: q=mcΔTq = m \cdot c \cdot \Delta T where:

    • mm is the mass of the solution (in grams)
    • cc is the specific heat capacity of the solution (assumed to be the same as water, 4.18J/gC4.18 \, \text{J/g}^\circ\text{C})
    • ΔT\Delta T is the change in temperature (TfinalTinitialT_{\text{final}} - T_{\text{initial}})
  2. Calculate the total volume and mass of the solution: Assuming the density of the solutions is approximately 1.00g/mL1.00 \, \text{g/mL}: Total volume=61.1mL+78.0mL=139.1mL\text{Total volume} = 61.1 \, \text{mL} + 78.0 \, \text{mL} = 139.1 \, \text{mL} Total mass=139.1g\text{Total mass} = 139.1 \, \text{g}

  3. Calculate the change in temperature: ΔT=29.00C22.90C=6.10C\Delta T = 29.00^\circ\text{C} - 22.90^\circ\text{C} = 6.10^\circ\text{C}

  4. Calculate the heat change (qq): q=139.1g×4.18J/gC×6.10Cq = 139.1 \, \text{g} \times 4.18 \, \text{J/g}^\circ\text{C} \times 6.10^\circ\text{C} q3544.89Jq \approx 3544.89 \, \text{J} Since 1kJ=1000J1 \, \text{kJ} = 1000 \, \text{J}, we convert qq to kJ: q3.54489kJq \approx 3.54489 \, \text{kJ}

  5. Determine the number of moles of HNO3\text{HNO}_3 reacted: Number of moles of HNO3=61.1mL×1.00mol1000mL=0.0611mol\text{Number of moles of } \text{HNO}_3 = 61.1 \, \text{mL} \times \frac{1.00 \, \text{mol}}{1000 \, \text{mL}} = 0.0611 \, \text{mol}

  6. Calculate the molar enthalpy of neutralization (ΔHneut\Delta H_{\text{neut}}): ΔHneut=qmoles of HNO3\Delta H_{\text{neut}} = \frac{q}{\text{moles of } \text{HNO}_3} ΔHneut=3.54489kJ0.0611mol58.03kJ/mol\Delta H_{\text{neut}} = \frac{3.54489 \, \text{kJ}}{0.0611 \, \text{mol}} \approx 58.03 \, \text{kJ/mol}

So, the molar enthalpy of neutralization of nitric acid is approximately 58.03kJ/mol\boxed{58.03} \, \text{kJ/mol}.

Would you like further details or have any questions?

Here are 5 questions you might want to ask:

  1. How do you convert temperature changes to heat energy in calorimetry?
  2. What assumptions are made in calorimetry calculations?
  3. How does specific heat capacity affect the heat change in a solution?
  4. Can you explain the steps to balance a neutralization reaction equation?
  5. What are some common sources of error in calorimetry experiments?

Tip: When performing calorimetry experiments, ensure that the calorimeter is well-insulated to minimize heat exchange with the surroundings for more accurate measurements.

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Math Problem Analysis

Mathematical Concepts

Calorimetry
Heat capacity
Thermochemistry
Moles

Formulas

q = m \cdot c \cdot \Delta T
\Delta H_{\text{neut}} = \frac{q}{\text{moles of } \text{HNO}_3}

Theorems

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Suitable Grade Level

High School