1. Which of the following explains why the bomb calorimeter exemplifies the First Law of Thermodynamics.
a. It requires the slow flow of heat energy and information is not lost.
b. The loss of heat energy from the sample becomes the gain in heat energy of the water bath
c. It operates under standard isobaric conditions
2. When 2.00 g of calcium metal reacts with 100.00 g of 1.00 M HCl in a coffee-cup calorimeter, the temperature rises from 18.1ºC to 66.8ºC. The heat capacity of the calorimeter apparatus is 131 J/K. Calculate ∆H (in kJ) for the reaction:
Ca(s) + 2HCl(aq) → CaCl_2(aq) + H_2(g)
Assume the specific heat of the solution is 4.18 J/K∙g.
1. The correct answer is b. The loss of heat energy from the sample becomes the gain in heat energy of the water bath. The bomb calorimeter exemplifies the First Law of Thermodynamics because it allows for the conservation of energy. In this case, the heat released by the reaction in the bomb calorimeter is transferred to the surrounding water bath, resulting in an increase in its temperature.
2. To calculate ΔH for the reaction, we need to use the equation:
ΔH = q / n
where q is the heat absorbed or released by the reaction, and n is the number of moles of the limiting reactant.
First, let's calculate the heat absorbed or released by the reaction using the equation:
q = m × c × ΔT
where m is the mass of the solution, c is the specific heat of the solution, and ΔT is the change in temperature.
Given:
Mass of solution (m) = 100.00 g
Specific heat of the solution (c) = 4.18 J/K·g
Change in temperature (ΔT) = 66.8ºC - 18.1ºC = 48.7ºC
q = (100.00 g) × (4.18 J/K·g) × (48.7ºC)
q = 20,017 J
Next, let's calculate the number of moles of the limiting reactant, HCl.
Given:
Mass of calcium (Ca) (molar mass = 40.08 g/mol) = 2.00 g
Molarity of HCl (M) = 1.00 M
Number of moles of HCl = (100.00 g) / (1.00 M) = 100.00 mol
From the balanced chemical equation:
1 mole of Ca reacts with 2 moles of HCl
Therefore, the number of moles of HCl reacting with 2.00 g of Ca = (2.00 g Ca) × (1 mol Ca / 40.08 g) × (2 mol HCl / 1 mol Ca) = 0.0996 mol
Finally, we can calculate ΔH:
ΔH = q / n
ΔH = 20,017 J / 0.0996 mol
ΔH ≈ 201,587 J/mol
Converting to kilojoules:
ΔH ≈ 201.59 kJ (rounded to two decimal places)
Therefore, ΔH for the reaction Ca(s) + 2HCl(aq) → CaCl2(aq) + H2(g) is approximately 201.59 kJ.
To answer question 1, we need to understand the First Law of Thermodynamics, which states that energy cannot be created or destroyed in an isolated system, but it can be transferred from one part of the system to another in the form of heat or work.
a. The bomb calorimeter exemplifies the First Law of Thermodynamics because it requires the slow flow of heat energy and information is not lost. This option does not provide a direct explanation of how the bomb calorimeter exemplifies the First Law of Thermodynamics, so it is not the correct answer.
b. The loss of heat energy from the sample becomes the gain in heat energy of the water bath. This option directly explains how the bomb calorimeter exemplifies the First Law of Thermodynamics. In a bomb calorimeter, the heat released by the reaction is absorbed by the surrounding water bath, obeying the principle of energy conservation. Therefore, this option is the correct answer.
c. It operates under standard isobaric conditions. This option refers to the operating conditions of the bomb calorimeter, but it does not directly explain how it exemplifies the First Law of Thermodynamics. Therefore, it is not the correct answer.
To answer question 2, we can use the equation:
q = m × c × ∆T
where q is the heat absorbed or released, m is the mass, c is the specific heat capacity, and ∆T is the change in temperature.
First, we need to calculate the heat released by the reaction, which is equal to the heat absorbed by the calorimeter:
q_calorimeter = -q_reaction
The negative sign indicates that the heat released by the reaction is taken as a negative value because it is being absorbed by the calorimeter.
First, let's calculate the heat absorbed or released by the calorimeter:
q_calorimeter = 131 J/K × (66.8ºC - 18.1ºC)
Next, we can calculate the heat absorbed or released by the reaction using the formula:
q_reaction = q_solution + q_calorimeter
The heat absorbed by the solution can be calculated as:
q_solution = m_solution × c_solution × ∆T
q_solution = (100.00 g + 2.00 g) × (4.18 J/gºC) × (66.8ºC - 18.1ºC)
Now, we can substitute these values into the equation for q_reaction:
q_reaction = (100.00 g + 2.00 g) × (4.18 J/gºC) × (66.8ºC - 18.1ºC) + 131 J/K × (66.8ºC - 18.1ºC)
Finally, we can convert the heat released by the reaction from J to kJ by dividing by 1000:
∆H_reaction = q_reaction / 1000
By evaluating these calculations, you should be able to determine the value of ∆H_reaction in kJ for the given reaction.
1. So what is the first law of thermodynamics? Apply that for the answer.
2.Total q = mass solution x sp.h x (Tfinal - Tinitial) + Ccal(Tfinal-Tinitial)
Than subtract the Ccal*dT.