Calculate the molar heat of neutralization in kJ/mol of the reaction between HA and BOH given the following information:
The temperature change equals 7C,
50 mL of 1 M concentration of Acid
50 mL of 1 M concentration of Base
Heat capacity of the calorimeter is 6.5 J/C.
Remember that the specific heat of water is 4.18 J/gC
q = (mass H2O x specific heat H2O x delta T) + (heat cap calorimeter x delta T).
q = (100 g x 4.18 x 7) + (6.5 x 7)= ?? joules.
Change to kJ. You want kJ/mol, you have kJ/0.05 mol.
To calculate the molar heat of neutralization, you need to determine the amount of heat released or absorbed during the reaction between HA and BOH.
Step 1: Calculate the heat transferred
First, calculate the heat transferred during the reaction using the formula:
q = m * c * ΔT
where,
q = heat transferred
m = mass of the solution (in grams)
c = specific heat of water (4.18 J/g·°C)
ΔT = change in temperature (in °C)
Since both the acid and base are 1 M concentration, we can assume their densities are approximately 1 g/mL.
However, since 50 mL of each solution is mixed, the total mass is:
m = (50 mL + 50 mL) * (1 g/mL) = 100 g
Using the given information, we have:
q = 100 g * 4.18 J/g·°C * 7 °C
q = 2926 J
Step 2: Convert the heat transferred to kilojoules (kJ)
To convert the heat transferred from joules to kilojoules, divide the value by 1000:
q = 2926 J ÷ 1000
q = 2.926 kJ
Step 3: Calculate the moles of the limiting reactant
To determine the moles of the limiting reactant, divide the volume of the solution by 1000 and then multiply it by the molarity.
For the acid, we have:
moles of HA = (50 mL ÷ 1000 mL) * 1 M
moles of HA = 0.05 mol
Step 4: Calculate the molar heat of neutralization
The molar heat of neutralization is calculated by dividing the heat transferred (in kilojoules) by the moles of the limiting reactant:
molar heat of neutralization = q ÷ moles of limiting reactant
molar heat of neutralization = 2.926 kJ ÷ 0.05 mol
molar heat of neutralization = 58.52 kJ/mol
Therefore, the molar heat of neutralization in kJ/mol of the reaction between HA and BOH is approximately 58.52 kJ/mol.
To calculate the molar heat of neutralization, we can use the formula:
q = m * c * ΔT
Where:
q = heat absorbed or released
m = mass of solution
c = specific heat capacity of water
ΔT = temperature change
First, we need to calculate the mass of the solution. Since we have 50 mL each of acid and base, the total volume of the solution will be 100 mL.
100 mL = 0.1 L (since 1 L = 1000 mL)
The concentration of the acid and base is 1 M, which means that there is 1 mole of acid and 1 mole of base in 1 liter of solution.
100 mL = 0.1 L = 0.1 mol
Now, let's calculate the heat change (q) using the given values:
q = (0.1 mol) * (4.18 J/g°C) * (7°C)
q = 2.93 J
Since the heat capacity of the calorimeter is 6.5 J/°C, we need to convert the heat change to match the calorimeter's heat capacity:
q = (2.93 J) / (6.5 J/°C)
q = 0.45 °C
To convert this to kilojoules per mole (kJ/mol), we need to calculate the molar heat of neutralization:
Molar heat of neutralization = (0.45 kJ) / (0.1 mol)
Molar heat of neutralization = 4.5 kJ/mol
Therefore, the molar heat of neutralization in kJ/mol for the reaction between HA and BOH is 4.5 kJ/mol.