A 100.0 ml sample of 0.300 M NaOH is mixed with a 100.0 ml sample of 0.300 M HNO3 in a coffee cup calorimeter. Both solutions were initially at 35.0 degrees celcius; the temperature of the resulting solution was recorded at 37.0 degrees celcius. Determine the delta H (in units of kJ/mol NaOH) for the neutralization reaction between aqueous NaOH and HNO3. Assume no heat is lost to the calorimeter or the surroundings; and the density and heat capacity of the resulting solution are the same as water.
What is delta H?
What is delta H/mol NaoH
To determine the delta H (enthalpy change) for the neutralization reaction between NaOH and HNO3, we can use the equation:
q = m * C * ΔT
where:
- q represents the heat transferred (in Joules)
- m represents the mass of the solution (in grams)
- C represents the heat capacity of the solution (in J/(g·°C))
- ΔT represents the change in temperature (in °C)
First, let's convert the volumes of the solutions into grams. We need to know the density and molar mass of the solutions for this conversion.
The density of water is approximately 1 g/mL. Since the density and heat capacity of the resulting solution are assumed to be the same as water, we can assume that 1 mL of solution is equivalent to 1 gram.
Next, we need to calculate the mass of NaOH and HNO3 in the solutions using their molar concentrations (M).
Molarity (M) = moles solute / liters solution
Moles solute = Molarity (M) * liters solution
For the NaOH solution:
Molarity (NaOH) = 0.300 M
Volume (NaOH) = 100.0 mL = 100.0 grams (since 1 mL = 1 gram)
Liters (NaOH) = 100.0 grams / 1000 = 0.1 L
Moles NaOH = 0.300 M * 0.1 L = 0.030 moles
For the HNO3 solution:
Molarity (HNO3) = 0.300 M
Volume (HNO3) = 100.0 mL = 100.0 grams (since 1 mL = 1 gram)
Liters (HNO3) = 100.0 grams / 1000 = 0.1 L
Moles HNO3 = 0.300 M * 0.1 L = 0.030 moles
Since the reaction between NaOH and HNO3 is a 1:1 neutralization reaction, the moles of NaOH and HNO3 are equal.
Now, we can calculate the heat transferred (q) using the formula:
q = m * C * ΔT
The given temperatures are:
Initial temperature = 35.0 °C
Final temperature = 37.0 °C
ΔT = Final temperature - Initial temperature = 37.0 °C - 35.0 °C = 2.0 °C
The heat capacity (C) of water is approximately 4.18 J/(g·°C).
q = (mass of solution) * (heat capacity of water) * (ΔT)
q = (0.1 g + 0.1 g) * (4.18 J/(g·°C)) * (2.0 °C)
q = 0.2 g * 4.18 J/(g·°C) * 2.0 °C
q = 1.672 J
Finally, to determine the delta H for the neutralization reaction (ΔH), we need to convert the heat transferred (q) from Joules to kJ/mol NaOH.
Since we have 0.030 moles of NaOH,
ΔH (in kJ/mol NaOH) = (heat transferred, q) / (moles of NaOH)
ΔH = 1.672 J / 0.030 moles = 55.73 J/mol
However, we need to convert J to kJ:
ΔH = 55.73 J/mol / 1000 = 0.05573 kJ/mol
Therefore, the delta H (in units of kJ/mol NaOH) for the neutralization reaction between NaOH and HNO3 is approximately 0.05573 kJ/mol NaOH.
Note: Always remember to consider the units and carefully follow the steps for calculations.