2) Calculate the Molar Enthalpy of Neutralization (ΔHn) in kJ/mol of the reaction between a monoprotic acid and a monoprotic base, given the following information:
The temperature change equals 5.06°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
This is how I calculated
5*5*5.06*6.50= 822.25/4.180= 196.7
You haven't used specific heat H2O anywhere.
q = mass H2O X specific heat H2O x delta T + Ccal*delta T.
What's the 5*5 bit?
100 mL solution = 100 g H2O
q = [100g x 4.180 x 5.06] + [6.50 x 5.06] = ?
q/n = dH/mol = ?/0.05 = x.
To calculate the molar enthalpy of neutralization (ΔHn), you need to use the equation:
ΔHn = q / n
where:
ΔHn is the molar enthalpy of neutralization in kJ/mol,
q is the heat absorbed or released during the reaction,
n is the number of moles of acid or base used in the reaction.
Let's break down the calculation step by step using the given information:
1. Calculate the heat absorbed or released (q) during the reaction using the equation:
q = m * c * ΔT
where:
m is the mass of water being heated or cooled (in grams),
c is the specific heat of water (4.180 J/g°C),
ΔT is the temperature change in Celsius.
In this case, since 50.0 mL of acid and 50.0 mL of base are used, the total volume of solution is 100.0 mL or 100.0 grams (assuming the density of the solutions is the same as water).
2. Calculate the mass of water (m) heated or cooled using the equation:
m = V * d
where:
V is the volume of water (100.0 mL or 100.0 grams),
d is the density of water (1.00 g/mL or g/cm³).
3. Calculate the temperature change (ΔT) in Celsius, which is given as 5.06°C.
4. Substitute the values of m, c, and ΔT into the equation q = m * c * ΔT to find the heat absorbed or released during the reaction.
q = 100.0 g * 4.180 J/g°C * 5.06°C = 2097.908 J
5. Calculate the number of moles of acid or base used (n) by using the equation:
n = c * V
where:
c is the concentration of the acid or base (1.00 M or mol/L),
V is the volume of the acid or base used (50.0 mL or 0.0500 L).
6. Substitute the values of c and V into the equation n = c * V to find the number of moles.
n = 1.00 mol/L * 0.0500 L = 0.0500 mol
7. Finally, plug the values of q and n into the equation ΔHn = q / n to calculate the molar enthalpy of neutralization (ΔHn) in kJ/mol.
ΔHn = 2097.908 J / 0.0500 mol = 41.95816 kJ/mol
Therefore, the molar enthalpy of neutralization (ΔHn) for the given reaction is approximately 41.96 kJ/mol.