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 evolved or absorbed during the reaction, and n is the number of moles of the limiting reactant.
Step 1: Calculate the heat evolved or absorbed (q)
To calculate q, we need to consider both the heat transferred to the solution and the heat transferred to the calorimeter.
The heat transferred to the solution can be calculated using the formula:
q_solution = mass × specific heat of water × ΔT
Given:
mass = volume × density = 50.0 mL × 1.00 g/mL = 50.0 g
specific heat of water = 4.180 J/g°C
ΔT = temperature change = 9.80 °C
So, q_solution = 50.0 g × 4.180 J/g°C × 9.80 °C = 2,038.8 J
The heat transferred to the calorimeter can be calculated using the formula:
q_calorimeter = heat capacity of calorimeter × ΔT
Given:
heat capacity of calorimeter = 6.50 J/°C
ΔT = temperature change = 9.80 °C
So, q_calorimeter = 6.50 J/°C × 9.80 °C = 63.7 J
The total heat transferred (q) is the sum of q_solution and q_calorimeter:
q = q_solution + q_calorimeter = 2,038.8 J + 63.7 J = 2,102.5 J
Step 2: Calculate the number of moles of the limiting reactant (n)
To calculate the number of moles of the limiting reactant, we need to use the concentration and volume of the acid or base.
Given:
volume of acid = 50.0 mL = 0.0500 L (since 1 mL = 0.001 L)
concentration of acid = 1.00 M
The number of moles of the limiting reactant can be calculated using the formula:
n = volume × concentration
So, n = 0.0500 L × 1.00 mol/L = 0.0500 mol
Step 3: Calculate the molar enthalpy of neutralization (ΔH)
Now, we can use the formula:
ΔH = q / n
Given:
q = 2,102.5 J
n = 0.0500 mol
ΔH = 2,102.5 J / 0.0500 mol = 42,050 J/mol
To convert J/mol to kJ/mol, divide by 1000:
ΔH = 42,050 J/mol / 1000 = 42.05 kJ/mol
So, the molar enthalpy of neutralization of the reaction is 42.05 kJ/mol.