I got an answer of -1.34 x 10^3 KJ/mol.
A 10.00 mL aliquot of 0.100 M NaOH is added to 15.00 mL of 0.100M HCl in a coffee cup calorimeter. The initial temperature of the two dilute liquids were both 25.0 degrees C and the final temperature of the mixture was found to be 37.8 degrees C. What is the (molar) heat of reaction for the neutralization?
Chemistry - DrBob222, Saturday, March 12, 2016 at 9:41pm
I suggest you check the numbers.
To calculate the molar heat of reaction for neutralization, you can use the formula:
q = m * C * ΔT
Where:
q = heat energy (in joules or kilojoules)
m = mass of the solution (in grams or kilograms)
C = specific heat capacity of the solution (in J/g·°C or kJ/kg·°C)
ΔT = change in temperature (in °C)
First, you need to calculate the mass of the solution. This can be done by adding the masses of the NaOH and HCl solutions.
mass of NaOH solution = volume of NaOH solution * density of NaOH solution
mass of NaOH solution = 10.00 ml * 1.00 g/ml = 10.00 g
mass of HCl solution = volume of HCl solution * density of HCl solution
mass of HCl solution = 15.00 ml * 1.00 g/ml = 15.00 g
Total mass of solution = mass of NaOH solution + mass of HCl solution
Total mass of solution = 10.00 g + 15.00 g = 25.00 g
Next, you need to calculate the change in temperature, ΔT.
ΔT = final temperature - initial temperature
ΔT = 37.8 °C - 25.0 °C = 12.8 °C
Now, you need to determine the specific heat capacity of the solution. Since the specific heat capacity of water is commonly used for dilute solutions, you can assume that the specific heat capacity of the solution is approximately 4.18 J/g·°C or 4.18 kJ/kg·°C.
Using the formula q = m * C * ΔT:
q = 25.00 g * 4.18 J/g·°C * 12.8 °C
q ≈ 1344 J
Since the heat of reaction is usually expressed in kilojoules per mole (kJ/mol), you need to convert the heat energy in joules to kilojoules by dividing by 1000.
q = 1344 J / 1000
q ≈ 1.344 kJ
Finally, to determine the molar heat of reaction, divide the heat energy by the number of moles of NaOH, which can be calculated using the molarity and volume of NaOH.
moles of NaOH = molarity of NaOH * volume of NaOH solution
moles of NaOH = 0.100 M * 10.00 mL / 1000 mL/L
moles of NaOH = 0.001 moles
molar heat of reaction = q / moles of NaOH
molar heat of reaction = 1.344 kJ / 0.001 moles
molar heat of reaction ≈ 1.34 x 10^3 kJ/mol
Therefore, the molar heat of reaction for the neutralization is approximately -1.34 x 10^3 kJ/mol.