In a coffee-cup calorimeter, 1 mol NaOH and 1 mol HCl initially at 18.5 o C (Cel
ID: 970719 • Letter: I
Question
In a coffee-cup calorimeter, 1 mol NaOH and 1 mol HCl initially at 18.5 oC (Celsius) are mixed in 100g of water to yield the following reaction:
NaOH + HCl Na+(aq) + Cl-(aq) + H2O(l)
After mixing the temperature rises to 107 oC. Calculate the change in enthalpy of this reaction.
Specific heat of the solution = 4.184 J/(g oC)
State your answer in kJ with 3 significant figures. Don't forget to enter the unit behind the numerical answer.
The molecular weight of NaOH is 40.0 g/mol, and the molecular weight of HCl is 36.5 g/mol.
H =
Explanation / Answer
NaOH + HCl Na+(aq) + Cl-(aq) + H2O(l)
mass of solution = 100+40+36.5 = 176.5 grams
heat released = m*s*DT
= 176.5*4.18*(107-18.5)
= 65.3 kj
enthalpy of reaction = - 65.3 kj/mol