A 100 mL sample of 0.300M NaOH is mixed with a 100mL sample of 0.305M HNOs in a coffee cup calorimeter (constant pressure). If both solutions were initially at 35 C and the temperature of the resulting solution was recorded as 37°C, determine the AHan (in units of KJ/mol of the limiting reactant). Assume that no heat is lost to the calorimeter or the surroundings. Assume that all of the solutions have specific heat capacity and density of 4.186 /g.°C and 1 g/mL respectively
To determine ∆H (AHan), we need to use the relationship:
∆H = q / n
where q is the heat absorbed or released by the reaction, and n is the number of moles of the limiting reactant.
In this case, we can consider the reaction between NaOH and HNO3 as a neutralization reaction, which produces NaNO3 and water.
1. Calculate the number of moles of NaOH and HNO3:
moles of NaOH = concentration × volume = 0.300 M × 0.100 L = 0.030 moles
moles of HNO3 = concentration × volume = 0.305 M × 0.100 L = 0.0305 moles
2. Determine the limiting reactant:
The limiting reactant is the one that is not in excess. Since the moles of NaOH and HNO3 are equal, both are the limiting reactants.
3. Calculate the heat exchanged by the reaction (q):
q = mc∆T
where m is the mass of the solution and c is the specific heat capacity.
First, calculate the mass of the solution:
mass = density × volume
mass = 1 g/mL × 0.100 L = 100 g
Second, calculate ∆T:
∆T = Tfinal - Tinitial
∆T = 37°C - 35°C = 2°C
Now, calculate q:
q = 100 g × 4.186 J/g°C × 2°C = 837.2 J
4. Convert q to kilojoules:
1 kJ = 1000 J
q = 837.2 J / 1000 = 0.8372 kJ
5. Calculate ∆H (AHan):
∆H = q / n
∆H = 0.8372 kJ / 0.030 moles = 27.907 kJ/mol
Therefore, the AHan (in units of kJ/mol of the limiting reactant) is 27.907 kJ/mol.