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In lab, you plan to carry out a calorimetry experiment to determine \(\Delta_{\mathrm{r}} H\) for the exothermic reaction of \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})\) and \(\mathrm{HCl}(\mathrm{aq}) .\) Predict how each of the following will affect the calculated value of \(\Delta_{\mathrm{r}} H\). (The value calculated for \(\Delta_{\mathrm{r}} H\) for this reaction is a negative value so choose your answer from the following: \(\Delta_{r} H\) will be too low [that is, a larger negative value], \(\Delta_{\mathrm{r}} H\) will be unaffected, \(\Delta_{\mathrm{r}} H\) will be too high \([\) that is, a smaller negative value. \(]\) )(a) You spill a little bit of the \(\mathrm{Ca}(\mathrm{OH})_{2}\) on the benchtop before adding it to the calorimeter. (b) Because of a miscalculation, you add an excess of HCl to the measured amount of \(\mathrm{Ca}(\mathrm{OH})_{2}\) in the calorimeter. (c) \(\mathrm{Ca}(\mathrm{OH})_{2}\) readily absorbs water from the air. The \(\mathrm{Ca}(\mathrm{OH})_{2}\) sample you weighed had been exposed to the air prior to weighing and had absorbed some water. (d) After weighing out \(\mathrm{Ca}(\mathrm{OH})_{2},\) the sample sat in an open beaker and absorbed water. (e) You delay too long in recording the final temperature. (f) The insulation in your coffee-cup calorimeter was poor, so some energy as heat was lost to the surroundings during the experiment. (g) You have ignored the fact that energy as heat also raised the temperature of the stirrer and the thermometer in your system.

Short Answer

Expert verified
(a) Too low; (b) Unaffected; (c) Too low; (d) Too low; (e) Too low; (f) Too high; (g) Too high.

Step by step solution

01

Analyze the effect of spilling reactant

When some of the \( \mathrm{Ca} ( \mathrm{OH} ) _{2} \) is spilled, less reactant is used in the reaction than intended. This means less heat is released than expected, leading to a calculated \( \Delta_r H \) that is too low (more negative).
02

Examine the effect of excess \( \mathrm{HCl} \)

Adding an excess of \( \mathrm{HCl} \) only affects the calculation if it affects the stoichiometry of the reaction. Since \( \Delta_r H \) is based on the limiting reactant, the calculated \( \Delta_r H \) will be unaffected as long as all \( \mathrm{Ca} ( \mathrm{OH} ) _{2} \) reacts.
03

Consider water absorption by \( \mathrm{Ca} ( \mathrm{OH} ) _{2} \)

\( \mathrm{Ca} ( \mathrm{OH} ) _{2} \) absorbing water affects the mass measurement, causing less than pure \( \mathrm{Ca} ( \mathrm{OH} ) _{2} \) to be used. This leads to less heat released than calculated, making \( \Delta_r H \) too low (more negative).
04

Impact of delay in temperature recording

Delaying the final temperature reading may allow heat to transfer to the surroundings, reducing the measured temperature change. This causes the calculated \( \Delta_r H \) to be too low (more negative) as less heat appears to be released.
05

Evaluate calorimeter insulation loss

Poor insulation results in heat loss to the surroundings. This makes the reaction seem less exothermic, leading to a calculated \( \Delta_r H \) that is too high (less negative).
06

Ignore heat absorbed by devices

If the heat absorbed by the stirrer and thermometer is ignored, the overall heat released seems lesser than it is. This results in a calculated \( \Delta_r H \) that is too high (less negative).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enthalpy Change
Enthalpy change, denoted as \(\Delta H\), is a fundamental concept in thermochemistry. It refers to the heat absorbed or released during a chemical reaction at constant pressure. For our specific reaction, the enthalpy change (\(\Delta_{\text{r}} H\)) is negative, indicating the reaction releases energy. This means we are dealing with an exothermic process.
When discussing enthalpy change, it is important to consider the effect of the reactants used in the experiment as well as the conditions under which the experiment is conducted. Inaccuracies such as spilling reactants or absorbing moisture can lead to incorrect values for \(\Delta_{\text{r}} H\). These errors alter the amount of substance that actually reacts, thus impacting the calculated heat change.
  • If not all of \(\mathrm{Ca} ( \mathrm{OH} ) _{2} \) is used due to spillage or water absorption, the calculated \(\Delta_{\text{r}} H\) might show a larger negative value (too low) because less heat is measured than should have been produced by the complete reaction.
  • Correct measurement and conduct during experiments highlight the importance of precision and attention to detail in laboratory practices.
Exothermic Reaction
Exothermic reactions release energy to their surroundings, typically in the form of heat. These reactions have \(\Delta H\) values that are negative. In our experiment with \(\mathrm{Ca} ( \mathrm{OH} ) _{2} \) and \(\mathrm{HCl}\), the exothermic nature means the reaction's products are of lower energy compared to the reactants.
Such energy release is the reason why the measured temperature within the calorimeter increases. However, various factors can obscure this temperature change, making it seem lower than expected, thus affecting our reading of \(\Delta_{\text{r}} H\).
  • Heat lost to the atmosphere due to poor insulation can reduce the apparent temperature change, skewing it to appear less exothermic (higher \(\Delta H\), less negative).
  • Factors like delay in measuring the temperature post-reaction can result in heat loss, inaccurately indicating a lower energy release.
Understanding these influences on an exothermic reaction is crucial to obtaining accurate experimental results.
Limiting Reactant
The limiting reactant is the substance that is entirely consumed when the chemical reaction completes, dictating the maximum extent of the reaction and thus the total energy change. In calorimetry, if the heat of reaction is calculated based on the limiting reactant, it's essential to correctly identify it for an accurate enthalpy change calculation.
In our reaction, if \(\mathrm{Ca} ( \mathrm{OH} ) _{2} \) is the limiting reactant, any excess of \(\mathrm{HCl}\) won’t change \(\Delta_{\text{r}} H\) as long as all \(\mathrm{Ca} ( \mathrm{OH} ) _{2} \) reacts. However:
  • Mis-identifying the limiting reactant or having accidental excess or missing amounts leads to incorrect heat calculation.
  • Proper stoichiometric calculations are vital to ensure that the measured enthalpy change corresponds genuinely to the theoretical reaction predictions.
Understanding the role of the limiting reactant can help in preempting potential experimental errors.
Heat Transfer
Heat transfer plays a significant role in calorimetry experiments. The goal is to ensure that most, if not all, the heat produced or consumed in a reaction is captured by the calorimeter. Any discrepancy in this can significantly affect the calculation of \(\Delta_{\text{r}} H\).
In our calorimetry setup, heat transfer inaccuracies could arise due to several factors:
  • Poor insulation means heat can escape into the surroundings, making the reaction appear less exothermic, which affects the \(\Delta_{\text{r}} H\) calculations by making them too high (less negative). "
  • If heat absorbed by additional components like the stirrer or thermometer isn't accounted for, the system records less heat than is actually released, again skewing the \(\Delta_{\text{r}} H\) to seem too high.
To mitigate such loss of heat and error in calculations, it's crucial to minimize external influences and accurately measure all aspects of heat transfer within the calorimeter.

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Most popular questions from this chapter

A 237 -g piece of molybdenum, initially at \(100.0^{\circ} \mathrm{C},\) was dropped into \(244 \mathrm{g}\) of water at \(10.0^{\circ} \mathrm{C} .\) When the system came to thermal equilibrium, the temperature was \(15.3^{\circ} \mathrm{C} .\) What is the specific heat capacity of molybdenum?

Three 45 -g ice cubes at \(0^{\circ} \mathrm{C}\) are dropped into \(5.00 \times 10^{2} \mathrm{mL}\) of tea to make iced tea. The tea was initially at \(20.0^{\circ} \mathrm{C} ;\) when thermal equilibrium was reached, the final temperature was \(0^{\circ} \mathrm{C} .\) How much of the ice melted, and how much remained floating in the beverage? Assume the specific heat capacity of tea is the same as that of pure water.

You have a large balloon containing 1.0 mol of gaseous water vapor at \(80^{\circ} \mathrm{C} .\) How will each step affect the internal energy of the system? (a) The temperature of the system is raised to \(90^{\circ} \mathrm{C}\) (b) The vapor is condensed to a liquid, at \(40^{\circ} \mathrm{C}\)

A 182 -g sample of gold at some temperature was added to 22.1 g of water. The initial water temperature was \(25.0^{\circ} \mathrm{C},\) and the final temperature was \(27.5^{\circ} \mathrm{C} .\) If the specific heat capacity of gold is \(0.128 \mathrm{J} / \mathrm{g} \cdot \mathrm{K},\) what was the initial temperature of the gold sample?

Insoluble \(\mathrm{PbBr}_{2}(\mathrm{s})\) precipitates when solutions of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})\) and \(\mathrm{NaBr}(\mathrm{aq})\) are mixed. \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+2 \mathrm{NaBr}(\mathrm{aq}) \rightarrow \mathrm{PbBr}_{2}(\mathrm{s})+2 \mathrm{NaNO}_{3}(\mathrm{aq})\) $$ \Delta_{\mathrm{r}} H^{\circ}=? $$ To measure the enthalpy change, \(200 .\) mL of \(0.75 \mathrm{M}\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})\) and \(200 . \mathrm{mL}\) of \(1.5 \mathrm{M} \mathrm{NaBr}(\mathrm{aq})\) are mixed in a coffee-cup calorimeter. The temperature of the mixture rises by \(2.44^{\circ} \mathrm{C} .\) Calculate the enthalpy change for the precipitation of \(\mathrm{PbBr}_{2}(\mathrm{s}),\) in \(\mathrm{k} \mathrm{J} / \mathrm{mol}\). (Assume the density of the solution is \(1.0 \mathrm{g} / \mathrm{mL},\) and its specific heat capacity is \(4.2 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}\).

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