Problem 67
Question
Suppose that you mix \(200.0 \mathrm{~mL}\) of \(0.200-\mathrm{M} \mathrm{RbOH}(\) aq \()\) with \(100 . \mathrm{mL}\) of \(0.400-\mathrm{M} \mathrm{HBr}(\mathrm{aq})\) in a coffee cup calorimeter. If the temperature of each of the two solutions was \(24.40^{\circ} \mathrm{C}\) before mixing, and the temperature rises to \(26.18^{\circ} \mathrm{C}\) (a) Calculate the heat transfer as a result of the reaction. (b) Write the thermochemical expression for the reaction.
Step-by-Step Solution
Verified Answer
The heat transfer is 2.23 kJ and the reaction is exothermic: \( \mathrm{RbOH(aq) + HBr(aq)} \rightarrow \mathrm{RbBr(aq) + H_2O(l)}; \Delta H = -2232.84\, \text{J/mol}\).
1Step 1: Determine moles of reactants
Calculate the moles of each reactant. \(0.200 \text{ L of } 0.200\, \mathrm{M} \mathrm{RbOH}\) gives \(0.200 \times 0.200 = 0.040\, \text{mol of } \mathrm{RbOH}\). \(0.100 \text{ L of } 0.400\, \text{M } \mathrm{HBr}\) gives \(0.100 \times 0.400 = 0.040\, \text{mol of } \mathrm{HBr}\). Since both reactants are present in equal moles and react in a 1:1 ratio, they will completely neutralize each other.
2Step 2: Calculate the mass of the solution
Assuming the densities are approximately that of water, the mass is the sum of the masses of both solutions. Density of water is \(1.00\, \text{g/mL}\), so total volume is \(200.0 + 100.0 = 300.0\, \text{mL}\). Thus, the mass is \(300.0\, \text{g}\).
3Step 3: Calculate heat absorbed by solution
Use the formula \(q = mc\Delta T\) where \(m = 300.0\, \text{g}\), \(c = 4.18\, \text{J/g°C (specific heat capacity of water)}\), and \(\Delta T = 26.18 - 24.40 = 1.78\, °C\). Thus \(q = 300.0 \times 4.18 \times 1.78 = 2232.84 \, \text{J}\) or \(2.23 \, \text{kJ}\).
4Step 4: Determine the thermochemical equation
The neutralization reaction is \(\mathrm{RbOH(aq)} + \mathrm{HBr(aq)} \rightarrow \mathrm{RbBr(aq)} + \mathrm{H_2O(l)} \). Since this is an exothermic reaction (the temperature rises), the enthalpy change is negative. The thermochemical equation is \(\mathrm{RbOH(aq)} + \mathrm{HBr(aq)} \rightarrow \mathrm{RbBr(aq)} + \mathrm{H_2O(l)} ; \Delta H = -2232.84 \, \text{J/mol}\) based on the total moles of the reaction occurring once with \(0.040\, \text{mol}\).
Key Concepts
Heat Transfer CalculationNeutralization ReactionThermochemical Equation
Heat Transfer Calculation
When two substances mix and react, they can absorb or release energy. This energy change is often measured in a process known as a heat transfer calculation. The formula to calculate this is
- \( q = mc\Delta T \),
- \( q \) is the amount of heat transferred,
- \( m \) is the mass of the solution,
- \( c \) is the specific heat capacity, and
- \( \Delta T \) is the change in temperature.
Neutralization Reaction
A neutralization reaction is a chemical reaction in which an acid and a base react to form water and a salt. This type of reaction generally happens when hydrogen ions from the acid combine with hydroxide ions from the base to produce water molecules.
In the exercise, we mixed equal moles of RbOH (the base) and HBr (the acid), both of which reacted completely to form RbBr and water, showing a typical neutralization. The initial and equal number of moles indicates that complete neutralization occurs, and no excess reactant remains. This specific type of combustion reaction is important in many applications, from industrial processes to everyday phenomena.
In the exercise, we mixed equal moles of RbOH (the base) and HBr (the acid), both of which reacted completely to form RbBr and water, showing a typical neutralization. The initial and equal number of moles indicates that complete neutralization occurs, and no excess reactant remains. This specific type of combustion reaction is important in many applications, from industrial processes to everyday phenomena.
Thermochemical Equation
Thermochemical equations combine chemical equations with the energy changes associated with a reaction. This tells us not only what substances you start with and end up with in a reaction, but also how the energy is changed in the process.
In our context, the thermochemical equation for the reaction between RbOH and HBr is:
In our context, the thermochemical equation for the reaction between RbOH and HBr is:
- \[ \mathrm{RbOH(aq)} + \mathrm{HBr(aq)} \rightarrow \mathrm{RbBr(aq)} + \mathrm{H_2O(l)} \] ;
- \( \Delta H = -2232.84 \, \text{J/mol} \),
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