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 concentrationof 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.
To calculate the molar enthalpy of neutralization, we need to use the equation:
ΔH = q / n
where ΔH is the molar enthalpy of neutralization, q is the heat absorbed or released in the reaction, and n is the number of moles of acid or base used in the reaction.
First, let's calculate the heat absorbed or released in the reaction using the formula:
q = mcΔT
where q is the heat, m is the mass of the solution, c is the specific heat capacity, and ΔT is the temperature change.
Since we have 50.0 mL of both the acid and base, the total volume of the solution will be 100.0 mL. To convert this to grams, we can use the density of water (1 g/mL) because both the acid and base concentrations are given.
Mass of the solution = 100.0 mL × 1 g/mL = 100.0 g
The specific heat capacity of water is 4.180 J/g°C, and the temperature change is given as 9.80 °C. So, plugging in the values, we get:
q = (100.0 g) × (4.180 J/g°C) × (9.80 °C) = 4089.84 J
Next, we need to calculate the number of moles of the acid used in the reaction. The concentration of the acid is given as 1.00 M, which means there is 1.00 mole of acid per 1.00 liter of solution. Since we have 50.0 mL of the acid solution, we can calculate the number of moles:
moles of acid = (50.0 mL) × (1.00 L/1000 mL) × (1.00 mol/L) = 0.0500 mol
Since the acid and base react in a 1:1 molar ratio, the number of moles of the base is also 0.0500 mol.
Finally, we can calculate the molar enthalpy of neutralization using the formula:
ΔH = q / n
ΔH = (4089.84 J) / (0.0500 mol) = 81796.8 J/mol
To convert this to kJ/mol, we divide by 1000:
ΔH = 81796.8 J/mol / 1000 = 81.7968 kJ/mol
Therefore, the molar enthalpy of neutralization in kJ/mol is approximately -81.80 kJ/mol (note the negative sign, indicating an exothermic reaction).