A mixture of NaOH and HCl is formed in a calorimeter. There is 50 mL of 2.078 Molar HCl and 51.7mL of 2.134 Molar NaOH.
How would the heat capacity of the calorimeter change if you were to double the volume of NaOH? Justify your answer
To determine how the heat capacity of the calorimeter would change if the volume of NaOH is doubled, we need to understand how the heat capacity is related to the volume of the solution.
The heat capacity of the calorimeter is a measure of its ability to absorb heat. It depends on the volume of the solution and the specific heat capacity of the solution itself.
When the volume of NaOH is doubled, the overall volume of the solution in the calorimeter will increase. As a result, the heat capacity of the calorimeter will also increase. This is because a larger volume of solution requires more heat energy to raise its temperature by the same amount.
Mathematically, the heat capacity (C) is given by the formula:
C = Q / ΔT
where Q is the heat absorbed or released and ΔT is the change in temperature. In this case, we are only considering the volume change and not the change in temperature.
Since the volume of the solution in the calorimeter has doubled, we can assume that the heat capacity of the calorimeter will also double proportionally.
Therefore, if you were to double the volume of NaOH in the mixture, the heat capacity of the calorimeter would also double.
Note: This assumes that the heat capacity of the solution remains the same and that there are no other factors affecting the heat capacity of the calorimeter.
To determine how the heat capacity of the calorimeter would change if you double the volume of NaOH, you need to understand how the heat capacity is related to the amount of heat absorbed or released during a chemical reaction.
First, let's determine the initial amount of heat released or absorbed during the reaction.
Since we have the volumes and concentrations of both HCl and NaOH, we can use the formula:
n = C x V
Where n represents the number of moles, C represents the concentration in molarity, and V represents the volume in liters.
For HCl:
n_HCl = C_HCl x V_HCl
= 2.078 mol/L x 0.050 L
= 0.1039 moles
Similarly, for NaOH:
n_NaOH = C_NaOH x V_NaOH
= 2.134 mol/L x 0.0517 L
= 0.1100 moles
Next, let's consider the balanced chemical equation for the reaction between NaOH and HCl, which produces water and sodium chloride:
NaOH + HCl -> NaCl + H2O
The balanced equation tells us that 1 mole of NaOH reacts with 1 mole of HCl to produce 1 mole of NaCl and 1 mole of water.
Since the reaction is a 1:1 ratio between NaOH and HCl, and the number of moles of each are both less than 1, it means that the reaction will go to completion. Thus, all of the NaOH and HCl will react with each other.
Now, let's calculate the amount of heat released or absorbed during the reaction using the balanced equation and the enthalpy of the reaction.
Assuming the reaction is exothermic (releases heat), the heat released (q) can be calculated using the formula:
q = n x ∆H
Where q represents the heat released or absorbed, n represents the number of moles, and ∆H represents the enthalpy change of the reaction.
In this case, ∆H is the enthalpy change for the reaction between NaOH and HCl, which is a known value.
For simplicity, let's assume that ∆H is -55.9 kJ/mole, which is the value for the reaction between NaOH and HCl.
For HCl:
q_HCl = n_HCl x ∆H
= 0.1039 moles x -55.9 kJ/mole
= -5.8 kJ
Similarly, for NaOH:
q_NaOH = n_NaOH x ∆H
= 0.1100 moles x -55.9 kJ/mole
= -6.1 kJ
Now, let's calculate the initial total heat released or absorbed by adding the heat values for HCl and NaOH:
q_initial = q_HCl + q_NaOH
= -5.8 kJ + (-6.1 kJ)
= -11.9 kJ
The negative sign indicates that the reaction releases heat.
Now, let's consider what would happen if we were to double the volume of NaOH.
If we double the volume of NaOH, the new volume would be 2 x 51.7mL = 103.4 mL.
To calculate the new amount of heat released or absorbed, we need to recalculate the number of moles for NaOH and then calculate the new heat values.
For the new volume of NaOH:
n_NaOH_new = C_NaOH x V_NaOH_new
= 2.134 mol/L x 0.1034 L
= 0.2203 moles
Now, let's calculate the new heat released or absorbed:
q_NaOH_new = n_NaOH_new x ∆H
= 0.2203 moles x -55.9 kJ/mole
= -12.3 kJ
Finally, let's calculate the new total heat released or absorbed by adding the new heat value for NaOH to the previous heat value for HCl:
q_new = q_HCl + q_NaOH_new
= -5.8 kJ + (-12.3 kJ)
= -18.1 kJ
Comparing the initial total heat of -11.9 kJ to the new total heat of -18.1 kJ, we can see that the heat capacity of the calorimeter would increase if we double the volume of NaOH. The negative sign in both cases indicates a heat release. However, the magnitude of the heat release increases from -11.9 kJ to -18.1 kJ, indicating that more heat is released in the larger reaction mixture. This signifies that the heat capacity of the calorimeter increases when the volume of NaOH is doubled.