Calculate the Molar Enthalpy of Neutralization (ÄHn) in kJ/mol of the reaction between a monoprotic acid and a monoprotic base, given the following information:
The temperature change equals 5.30°C, 50.0 mL of 1.00 M concentration of Acid 50.0 mL of 1.00 M concentration of Base Heat capacity of the calorimeter is 6.50 J/°C. The specific heat of water is 4.180 J/g°C
To calculate the molar enthalpy of neutralization (ΔHn) in kJ/mol, we need to follow a few steps:
Step 1: Determine the moles of acid and base used:
To find the moles of acid and base used, we need to use the given volume and concentration. The formula we'll use is:
moles = volume (L) × concentration (mol/L)
Since the volumes of both the acid and base are given as 50.0 mL (0.050 L) and the concentration is given as 1.00 M, the number of moles for both the acid and base will be:
moles = 0.050 L × 1.00 mol/L = 0.050 mol
Step 2: Calculate the heat transferred:
The heat transferred during the neutralization process can be calculated using the formula:
q = m × c × ΔT
where q is the heat energy transferred, m is the mass of the solution, c is the specific heat capacity, and ΔT is the temperature change.
In this case, we must calculate the mass of the solution, which is the combined mass of the acid and base solution. Since both volumes are 50.0 mL (0.050 L), and the density of water is approximately 1 g/mL, the mass of the solution will be:
mass of solution = volume × density = 0.050 L × 1 g/mL = 0.050 g
Now we can calculate the actual heat transferred:
q = (mass of solution) × (specific heat capacity of water) × (temperature change)
q = 0.050 g × 4.180 J/g°C × 5.30°C
Step 3: Convert heat to kJ:
The answer obtained in the previous step is in joules. To convert it to kJ, we must divide the answer by 1000:
q (in kJ) = q (in J) / 1000
Now you have the value for q in kJ.
Step 4: Calculate the molar enthalpy (ΔHn):
To calculate the molar enthalpy of neutralization (ΔHn) in kJ/mol, we use the formula:
ΔHn = q (in kJ) / moles of acid or base used
Since the acid and base react in a 1:1 stoichiometric ratio, we can use moles of the acid or base interchangeably. Therefore, ΔHn will be:
ΔHn = q (in kJ) / 0.050 mol
Calculate the final value to obtain the molar enthalpy of neutralization (ΔHn) in kJ/mol.