10mL of 0.500M HCl was added to 50mL of 0.100M NaOH in a coffee cup calorimeter, and the temperature increased from 22.3°C to 23.1°C. What is the experimentally determined enthalpy of neutralization of HCl (in kJ/mol)? The specific heat of water is 4.184J/g•°C, and the heat capacity of the calorimeter is 10J/°C.
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To determine the enthalpy of neutralization (ΔH) for the reaction between HCl and NaOH, we can use the equation:
ΔH = q / n
where:
- ΔH is the enthalpy change (in kJ/mol),
- q is the heat transferred (in kJ),
- and n is the number of moles of HCl.
To find the heat transferred (q), we can use the equation:
q = mcΔT
where:
- q is the heat transferred (in kJ),
- m is the mass of the solution (in g),
- c is the specific heat of water (in J/g•°C),
- and ΔT is the change in temperature (in °C).
First, let's calculate the mass of the solution by summing the masses of the HCl and NaOH:
m(HCl) = 10 mL x 1.00 g/mL = 10.0 g
m(NaOH) = 50 mL x 1.00 g/mL = 50.0 g
m(solution) = m(HCl) + m(NaOH) = 10.0 g + 50.0 g = 60.0 g
Next, calculate the heat transferred (q) using the above equation:
q = mcΔT
q = (60.0 g) x (4.184 J/g•°C) x (23.1°C - 22.3°C) = 358.27 J
Now, we need to convert the heat transferred (q) to kilojoules (kJ):
q = 358.27 J / 1000 = 0.35827 kJ
Finally, we can determine the enthalpy of neutralization (ΔH) using:
ΔH = q / n
To find the number of moles of HCl (n), we can use the equation:
n = c x V
where:
- c is the concentration of HCl (in mol/L),
- and V is the volume of HCl (in L).
First, convert the volume of HCl (10 mL) to liters:
V(HCl) = 10 mL / 1000 = 0.01 L
Using the equation n = c x V, where the concentration of HCl (c) is 0.500M:
n = 0.500 mol/L x 0.01 L = 0.005 mol
Now we can calculate ΔH:
ΔH = q / n
ΔH = 0.35827 kJ / 0.005 mol = 71.654 kJ/mol
Therefore, the experimentally determined enthalpy of neutralization of HCl is 71.654 kJ/mol.