Calculate the Molar Enthalpy of Neutralization in kJ/mol of the reaction between a monoprotic acid and a monoprotic base, given the following information: *Make sure you report your answer using the correct sign*
The temperature change equals 9.80 degrees C, 50.0mL of 1.00 M concentration of Acid
50.0mL of 1.00 M concentration of Base, heat capacity of the calorimeter is 6.50 J/degree C. The specific heat of
water is 4.180 J/g degree C.
See my comments to your other post regarding this being a made up problem.
I don't like this answer either based on the same rationale as I posted earlier; however,
q = qH2O + qCal
q = [mass H2O x specific heat H2O x dT) + (CCal*dT)
q = (100.0 x 4.18 x 9.80)+(6.50*9.80) = 4160.1 J
q/mol = -4160.1 J/0.050 mol = -83.2 J/mol which is -0.0832 kJ/mol. This is a ridiculous answer based on literature values of 57.1 kJ/mol.
Again, please let me know how this turns out. If I'm making an error I need to know it.
To calculate the molar enthalpy of neutralization, we need to use the formula:
ΔH = q / n
Where:
ΔH is the molar enthalpy of neutralization (in kJ/mol),
q is the heat absorbed or released by the reaction (in J),
and n is the number of moles of the limiting reactant.
In this case, the limiting reactant is either the acid or the base, since they react in a 1:1 ratio.
First, let's calculate the heat absorbed or released by the reaction (q).
q = q_acid + q_base + q_water
1. Heat absorbed or released by the acid (q_acid):
q_acid = (m_acid) x (C_acid) x (ΔT_acid)
Where:
m_acid is the mass of the acid (in grams),
C_acid is the specific heat of the acid (in J/g°C),
and ΔT_acid is the change in temperature of the acid (in °C).
To find the mass of the acid, we can use its volume (50.0 mL) and concentration (1.00 M):
m_acid = (V_acid) x (C_acid) x (Molar mass of the acid)
The molar mass of the acid is needed to convert from moles to grams.
2. Heat absorbed or released by the base (q_base):
q_base = (m_base) x (C_base) x (ΔT_base)
Similarly, we can find the mass of the base using its volume (50.0 mL) and concentration (1.00 M):
m_base = (V_base) x (C_base) x (Molar mass of the base)
3. Heat absorbed or released by the water (q_water):
q_water = (m_water) x (C_water) x (ΔT_water)
Here, the mass of the water can be calculated using its density (1 g/mL) and volume (50.0 mL):
m_water = (V_water) x (Density of water)
Now that we have calculated q, we need to determine the number of moles of the limiting reactant (n). Since the acid and base are in a 1:1 ratio, the moles of acid will also be equal to the moles of base.
n = (V_acid) x (C_acid)
Finally, we can substitute the values into the formula:
ΔH = q / n
and convert the answer to kJ/mol.
Note: Always be cautious with units when performing calculations.