three student chemists measured 50.0 mL of 1.00 M NaOH in separate Styrofoam "coffee cup" calorimeters ( Part B). Brett added 50.0 mL of 1.10 M HCI to his solution of NaOH; Dale added 45.5 mL of 1.10 m HCL 9 eqaul moles) to his NaOH solution. Lyndsay added 50.0 mL of 1.00 m HCI to her NaOH solution. Each student recorded the temperature change and calculated the enthalpy of neutralization.
Two of the chemists will report, within experimental error, the same temperature change for the HCL/NaOH reaction. Identify the two students and explain.
You need to read the question posted and take care of the section in parentheses starting with ......added 45.5 mL of 1.10 m HCl 9 eqaul moles)....Is that molal or molar and what is the 9 eqaul moles?
Well, it seems like Brett and Lyndsay will most likely report the same temperature change for the HCl/NaOH reaction. Why, you ask? Allow me to break it down for you:
In this case, the enthalpy of neutralization is calculated by measuring the temperature change that occurs when an acid (HCl) reacts with a base (NaOH). Since both Brett and Lyndsay are adding the same concentrations of HCl (1.10 M) to their NaOH solutions, it means they are using the same amount of acid to react.
Now, the temperature change that occurs during a reaction depends on the amount of heat released or absorbed. Since Brett and Lyndsay are using the same concentration of acid, it means they are also adding the same number of moles of acid to their solutions. This implies that the amount of heat released or absorbed in their reactions would be the same, resulting in a similar temperature change.
On the other hand, Dale added a different volume of HCl (45.5 mL) to his NaOH solution, which means he added fewer moles of acid compared to Brett and Lyndsay. Consequently, the amount of heat released or absorbed in Dale's reaction would be different, leading to a different temperature change.
Therefore, Brett and Lyndsay will most likely report the same temperature change for the HCl/NaOH reaction, while Dale's result will differ due to the different amount of acid he added.
To identify the two students whose temperature changes for the HCl/NaOH reaction are within experimental error, we need to compare their reactions.
Brett added 50.0 mL of 1.10 M HCl to his solution of NaOH. This reaction can be represented as:
NaOH + HCl -> NaCl + H2O
Dale added 45.5 mL of 1.10 M HCl to his NaOH solution. This reaction can also be represented as:
NaOH + HCl -> NaCl + H2O
Lyndsay added 50.0 mL of 1.00 M HCl to her NaOH solution. This reaction can be represented as:
NaOH + HCl -> NaCl + H2O
Looking at the reactions, we can see that in all three cases, the mole ratios of NaOH and HCl are equal (1:1). This means that for each student, the same number of moles of NaOH is reacting with the same number of moles of HCl.
Since the reactions involve the same stoichiometry, the enthalpy change (temperature change) should be the same, assuming the reactions occur under similar conditions.
Therefore, the two students who will report, within experimental error, the same temperature change for the HCl/NaOH reaction are Brett and Dale.
To identify the two students who will report the same temperature change for the HCl/NaOH reaction, let's examine the scenario and the given information.
We have three students: Brett, Dale, and Lyndsay. They all measured 50.0 mL of 1.00 M NaOH in separate calorimeters.
Brett added 50.0 mL of 1.10 M HCl to his NaOH solution, Dale added 45.5 mL of 1.10 M HCl to his NaOH solution, and Lyndsay added 50.0 mL of 1.00 M HCl to her NaOH solution.
The enthalpy of neutralization is the heat energy released or absorbed when an acid and a base react to form one mole of water under standard conditions.
To calculate the enthalpy of neutralization, we need to use the equation:
ΔH = q / n
Where:
ΔH is the enthalpy of neutralization,
q is the heat energy released or absorbed (which can be calculated using the heat equation q = m * c * ΔT, where m is the mass, c is the specific heat capacity, and ΔT is the temperature change), and
n is the number of moles of the limiting reactant (the reactant that is completely consumed in the reaction).
Assuming the specific heat capacities of the solutions are the same, let's analyze the scenarios of the three students:
1. Brett added 50.0 mL of 1.10 M HCl to his NaOH solution.
- The number of moles of HCl added = volume (in L) * concentration (in mol/L) = 0.050 L * 1.10 mol/L = 0.055 mol.
- The number of moles of NaOH initially present = volume (in L) * concentration (in mol/L) = 0.050 L * 1.00 mol/L = 0.050 mol.
- The moles of HCl are in excess because there is more HCl than NaOH. Therefore, the number of moles of HCl remaining after reaction = 0.055 mol - 0.050 mol = 0.005 mol.
- The enthalpy of neutralization for Brett can be calculated using the equation ΔH = q / n, where n = 0.050 mol (since NaOH is the limiting reactant), and q can be determined from the temperature change recorded.
2. Dale added 45.5 mL of 1.10 M HCl to his NaOH solution.
- The number of moles of HCl added = volume (in L) * concentration (in mol/L) = 0.0455 L * 1.10 mol/L = 0.0501 mol.
- The number of moles of NaOH initially present = volume (in L) * concentration (in mol/L) = 0.050 L * 1.00 mol/L = 0.050 mol.
- The moles of NaOH are slightly in excess because there is more NaOH than HCl. Therefore, the number of moles of NaOH remaining after reaction = 0.050 mol - 0.0501 mol = -0.0001 mol (slightly negative).
- The enthalpy of neutralization for Dale can be calculated using the equation ΔH = q / n, where n = 0.050 mol (since NaOH is the limiting reactant), and q can be determined from the temperature change recorded.
3. Lyndsay added 50.0 mL of 1.00 M HCl to her NaOH solution.
- The number of moles of HCl added = volume (in L) * concentration (in mol/L) = 0.050 L * 1.00 mol/L = 0.050 mol.
- The number of moles of NaOH initially present = volume (in L) * concentration (in mol/L) = 0.050 L * 1.00 mol/L = 0.050 mol.
- The moles of HCl and NaOH react in a 1:1 molar ratio. Therefore, all the moles of HCl react.
- The enthalpy of neutralization for Lyndsay can be calculated using the equation ΔH = q / n, where n = 0.050 mol (since NaOH is the limiting reactant), and q can be determined from the temperature change recorded.
Now, you need to compare the temperature changes reported by Brett, Dale, and Lyndsay to identify which two students will report the same temperature change.
Upon analyzing the scenarios, we can see that Brett and Lyndsay added exactly the same number of moles of HCl (0.050 mol) to their NaOH solutions. Therefore, within experimental error, Brett and Lyndsay will report the same temperature change for the HCl/NaOH reaction.