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Chapter 5: Problem 115

A quantity of \(50.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) is mixed with \(50.0 \mathrm{~mL}\) of \(0.400 \mathrm{M} \mathrm{HNO}_{3}\) in a constant-pressure calorimeter having a heat capacity of \(496 \mathrm{~J} /{ }^{\circ} \mathrm{C}\). The initial temperature of both solutions is the same at \(22.4^{\circ} \mathrm{C}\). What is the final temperature of the mixed solution? Assume that the specific heat of the solutions is the same as that of water and the molar heat of neutralization is \(-56.2 \mathrm{~kJ} / \mathrm{mol}\).

Short Answer

Expert verified
The final temperature of the mixed solution is approximately \(23.6^{\circ} \text{C}\).

Step by step solution

01

Calculate Moles of Reactants

We need to calculate the moles of each reactant to determine how much of each is involved in the reaction. The moles of \(\text{Ba(OH)}_2\) are given by:\[\text{moles of } \text{Ba(OH)}_2 = 0.200 \text{ M} \times 0.050 \text{ L} = 0.010 \text{ moles}\]For \(\text{HNO}_3\), we have:\[\text{moles of } \text{HNO}_3 = 0.400 \text{ M} \times 0.050 \text{ L} = 0.020 \text{ moles}\]
02

Determine Limiting Reactant

In the reaction, \(\text{Ba(OH)}_2\) and \(\text{HNO}_3\) react in a 1:2 ratio:\[\text{Ba(OH)}_2 + 2 \text{HNO}_3 \rightarrow \text{Ba(NO}_3)_2 + 2 \text{H}_2\text{O}\]0.010 moles of \(\text{Ba(OH)}_2\) would need 0.020 moles of \(\text{HNO}_3\), which is available. Hence, both reactants are completely consumed without any limiting factor.
03

Calculate Heat Released

The reaction releases heat based on the moles of water formed (neutralized). Since 2 water molecules are formed per formula unit of \(\text{Ba(OH)}_2\), the reaction will produce:\[\text{moles of water} = 0.020 \text{ moles}\]Thus, the heat released is:\[q_{ ext{reaction}} = 0.020 \text{ moles} \times (-56.2 \text{ kJ/mol}) = -1.124 \text{ kJ} = -1124 \text{ J}\](The negative sign indicates exothermic reaction.)
04

Calculate Temperature Change

The heat absorbed by the calorimeter and the resulting solution is equivalent to the magnitude of the heat released during the reaction:\[q_{ ext{solution}} = -q_{ ext{reaction}} = 1124 \text{ J}\]Using the formula \(q = mc\Delta T + C_{cal}\Delta T\), where:- \(m\) is the mass of the solution (approximately equal to 100 g)- \(c\) is the specific heat of water \(\approx 4.18 \text{ J/g 掳C}\)- \(C_{cal}\) is the calorimeter's heat capacity \(496 \text{ J/掳C}\)- \(\Delta T\) is the temperature changeSubstitute into the equation to find \(\Delta T\):\[1124 = (100 \times 4.18 + 496) \Delta T \Rightarrow 1124 = 914 \times \Delta T\]Solve for \(\Delta T\):\[\Delta T = \frac{1124}{914} \approx 1.23 \text{ 掳C}\]
05

Determine Final Temperature

The initial temperature was \(22.4^{\circ} \mathrm{C}\). Add the temperature change to the initial temperature:\[T_{\text{final}} = 22.4^{\circ} \text{C} + 1.23^{\circ} \text{C} = 23.63^{\circ} \text{C}\]

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Key Concepts

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

Heat of Neutralization
In chemistry, the heat of neutralization refers to the energy change when an acid and a base neutralize each other to form water and a salt. This reaction generally releases heat, making it exothermic. The process we see in the given reaction involves the neutralization of barium hydroxide, an inorganic base, with nitric acid, producing barium nitrate and water.
The heat of neutralization is standardized to measure how much energy is released or absorbed per mole of formation. For water formation in reactions involving strong acids and bases, the heat of neutralization is often around \(-56.2 \, \mathrm{kJ/mol}\), as is the case in our exercise. This means each mole of water formed releases \(56.2 \, \mathrm{kJ}\) of energy, signaling an exothermic process.
  • This standard heat is derived under constant pressure.
  • The reaction releases heat because the formation of water from H鈦� and OH鈦� ions is a relatively efficient energy release activity.
Limiting Reactant
In any chemical reaction, the limiting reactant is the substance that is totally consumed when the chemical reaction is complete, limiting the amount of product formed.

To determine the limiting reactant, we perform stoichiometric calculations.
  • The reaction between Ba(OH)鈧� and HNO鈧� is in a 1:2 mole ratio as expressed in the balanced chemical equation.
In our problem, calculations show that both reactants are consumed completely based on their mole ratio requirements:
  • 0.010 moles of Ba(OH)鈧� needs 0.020 moles of HNO鈧�.
  • Since the available HNO鈧� is exactly 0.020 moles, both reactants are completely consumed with no excess.
Thus, neither Ba(OH)鈧� nor HNO鈧� is in excess, meaning no limiting reactant is present as both are fully consumed.
Exothermic Reaction
An exothermic reaction is a chemical reaction that releases energy by light or heat. It's characterized by a negative change in enthalpy \(\Delta H\) < 0.
  • In the context of our exercise, the heat of neutralization being negative \(-56.2 \, \mathrm{kJ/mol}\) confirms the exothermic nature.
  • The heat released during one complete reaction is determined to be \(-1124 \, \mathrm{J}\).
As the reaction proceeds, the system loses heat to the surroundings, raising the temperature of the solution in the calorimeter. This increase is manageable to calculate, providing valuable insights into energy transfers during chemical reactions.

Understanding exothermic reactions is key in calorimetry, helping us measure how much energy is released during various reactions.
Specific Heat Capacity
Specific heat capacity is crucial when studying calorimetry. It reflects the amount of heat required to change a substance's temperature by one degree Celsius.

For this exercise, the specific heat capacity of the solutions is assumed to be the same as water, \(4.18 \, \mathrm{J/g掳C}\).
  • This assumption helps simplify calculations as water's specific heat is a well-documented standard.
  • Combining this with the calorimeter's heat capacity \(496 \, \mathrm{J/掳C}\) allows for accurate \(\Delta T\) computations.
The equation \(q = mc\Delta T + C_{cal}\Delta T\) incorporates these elements to calculate temperature changes. Hence, any alteration in temperature helps quantify energy exchange during the chemical reaction. By understanding the specific heat capacity, we ensure the precision of temperature prediction and energy balance calculations in calorimetry.

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Most popular questions from this chapter

Given the thermochemical data,$$\begin{array}{ll}\mathrm{A}+6 \mathrm{~B} \longrightarrow 4 \mathrm{C} & \Delta H_{1}=-1200 \mathrm{~kJ} / \mathrm{mol} \\\\\mathrm{C}+\mathrm{B} \longrightarrow \mathrm{D} & \Delta H_{1}=-150 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ Determine the enthalpy change for each of the following: a) \(\mathrm{D} \longrightarrow \mathrm{C}+\mathrm{B}\) d) \(2 \mathrm{D} \longrightarrow 2 \mathrm{C}+2 \mathrm{~B}\) b) \(2 \mathrm{C} \longrightarrow \frac{1}{2} \mathrm{~A}+3 \mathrm{~B}\) e) \(6 \mathrm{D}+\mathrm{A} \longrightarrow 10 \mathrm{C}\) c) \(3 \mathrm{D}+\frac{1}{2} \mathrm{~A} \stackrel{\longrightarrow}{\longrightarrow} \mathrm{C}\)

In writing thermochemical equations, why is it important to indicate the physical state (i.e., gaseous, liquid, solid, or aqueous) of each substance?

Calculate the amount of heat liberated (in kJ) from 366 \(\mathrm{g}\) of mercury when it cools from \(77.0^{\circ} \mathrm{C}\) to \(12.0^{\circ} \mathrm{C}\).

Describe the interconversions of forms of energy occurring in these processes: (a) You throw a softball up into the air and catch it. (b) You switch on a flashlight. (c) You ride the ski lift to the top of the hill and then ski down. (d) You strike a match and let it burn completely.

The standard enthalpy change \(\Delta H^{\circ}\) for the thermal decomposition of silver nitrate according to the following equation is \(+78.67 \mathrm{~kJ}\) : \(\mathrm{AgNO}_{3}(s) \longrightarrow \mathrm{AgNO}_{2}(s)+\frac{1}{2} \mathrm{O}_{2}(g)\) The standard enthalpy of formation of \(\mathrm{AgNO}_{3}(s)\) is \(-123.02 \mathrm{~kJ} / \mathrm{mol} .\) Calculate the standard enthalpy of formation of \(\mathrm{AgNO}_{2}(s)\).
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