Calculate the molar enthalpy change for this reaction:
HCl(aq 1.00M) + NaOH -> NaCl(aq,.500M)+ H2O
Initial temp: 22.15 degrees Celsius
Extrapolated temp: 25.87 degrees Celsius
DT: 3.72 degrees Celsius
Notes:
Calculate the enthalpy change for this reaction. The density of .500 M Nacl(aq) is 1.02 g/mL and its specific heat capacity is 4.02 J/g degree celsius.
No matter how long I stare at it, and look for videos online, I can not find a definite way to solve this.
As I see it you haven't listed a volume; i.e., only mols/L. Without a volume you can't calculate q.
50ml
To calculate the molar enthalpy change for the given reaction, we can use the equation:
q = m × C × ΔT
Where:
q is the heat absorbed or released during the reaction,
m is the mass of the solution (NaCl(aq)),
C is the specific heat capacity of the solution,
ΔT is the change in temperature.
First, we need to determine the mass of the NaCl(aq) solution. You mentioned the density is 1.02 g/mL. Since the concentration is 0.500 M, it means there are 0.500 moles of NaCl dissolved in 1 liter of solution. We can calculate the mass using the density:
mass = volume × density
mass = 1 L × 1.02 g/mL
mass = 1020 grams
Now that we have the mass of the solution, we can calculate the heat absorbed or released using the equation above.
q = m × C × ΔT
q = 1020 g × 4.02 J/g degrees Celsius × 3.72 degrees Celsius
q ≈ 15199 J
The heat change (q) corresponds to the enthalpy change (ΔH) for the reaction. However, this value is in joules, and we need to convert it to kilojoules per mole for the molar enthalpy change.
To convert from joules to kilojoules:
1 kJ = 1000 J
Hence,
ΔH = q / (1000 J/kJ)
ΔH = 15199 J / 1000 J/kJ
ΔH ≈ 15.199 kJ
Therefore, the molar enthalpy change for the reaction is approximately 15.199 kJ/mol.