Answer:
-62.12kJ/mol is heat of neutralization
Step-by-step explanation:
The neutralization reaction of HCl and NaOH is:
HCl + NaOH → NaCl + H₂O + HEAT
An acid that reacts with a base producing a salt and water
You can find the released heat of the reaction -heat of neutralization- (Released heat per mole of reaction) using the formula:
Q = C×m×ΔT
Where Q is heat, C specific heat of the solution (4.06J/gºC), m its mass of the solution and ΔT change in temperature (27.5ºC-20.0ºC = 7.5ºC).
The mass of the solution can be found with the volume of the solution (50.0mL of HCl solution + 50.0mL of NaOH solution = 100.0mL) and its density (1.02g/mL), as follows:
100.0mL × (1.02g / mL) = 102g of solution.
Replacing, heat produced in the reaction was:
Q = C×m×ΔT
Q = 4.06J/gºC×102g×7.5ºC
Q = 3106J = 3.106kJ of heat are released.
There are 50.0mL ×1M = 50.0mmoles = 0.0500 moles of HCl and NaOH that reacts releasing 3.106kJ of heat. That means heat of neutralization is:
3.106kJ / 0.0500mol of reaction =
-62.12kJ/mol is heat of neutralization
The - is because heat is released, absorbed heat has a + sign