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?
I suggest you check the numbers.
To find the molar heat of reaction for the neutralization, you can use the equation:
q = m * c * ΔT
where:
q is the heat transferred in joules (J)
m is the mass of the solution in grams (g)
c is the specific heat capacity of the solution in joules per gram per degree Celsius (J/g·°C)
ΔT is the change in temperature (final temperature - initial temperature) in degrees Celsius (°C)
However, in order to use this equation, we need to know the mass of the solution. To determine the mass of the solution, we can calculate it using the following equation:
mass = volume * density
Since we know the volume of the solution is the sum of the volumes of NaOH and HCl (10.00 mL + 15.00 mL = 25.00 mL), and the density of water is approximately 1.00 g/mL, we can calculate the mass of the solution.
mass = 25.00 mL * 1.00 g/mL
mass = 25.00 g
Now, let's calculate the heat transferred (q) using the given information. The specific heat capacity of the solution, c, is assumed to be the same as water, which is approximately 4.18 J/g·°C.
q = m * c * ΔT
q = 25.00 g * 4.18 J/g·°C * (37.8°C - 25.0°C)
q = 25.00 g * 4.18 J/g·°C * 12.8°C
q = 12897.60 J
Since the unit of q is in joules, we need to convert it into kilojoules (kJ) by dividing by 1000.
q = 12897.60 J / 1000
q = 12.89 kJ
Finally, to find the molar heat of reaction, we need to divide q by the moles of the limiting reactant. In this case, NaOH and HCl react in a 1:1 ratio, so the moles of NaOH is equal to the moles of HCl. To find the moles of HCl, we can use the equation:
moles = concentration * volume
moles HCl = 0.100 M * 15.00 mL
moles HCl = 0.00150 mol
Since the moles of NaOH and HCl are the same, the moles of NaOH is also 0.00150 mol.
Now, we can find the molar heat of reaction:
ΔH = q / moles NaOH
ΔH = 12.89 kJ / 0.00150 mol
ΔH ≈ -8593.33 kJ/mol
The negative sign indicates that the reaction is exothermic, meaning that heat is released. Therefore, the molar heat of reaction for the neutralization is approximately -8593.33 kJ/mol.