In a coffee-cup calorimeter, 1 mol KOH and 1 mol HBr initially at 27.5 o C (Cels
ID: 831298 • Letter: I
Question
In a coffee-cup calorimeter, 1 mol KOH and 1 mol HBr initially at 27.5 oC (Celsius) are mixed in 100g of water to yield the following reaction:
KOH + HBr ? K+(aq) + Br-(aq) + H2O(l)
After mixing the temperature rises to 82.2 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 KOH is 56.2 g/mol, and the molecular weight of HBr is 80.9 g/mol.
Explanation / Answer
Since there is 1mol of KOH and 1mol of HBr, you know that 1mol of KOH is equal to 56.2g and 1mol of HBr is equal to 80.9g. Now add those masses with the mass of water to get the total mass.
(56.2+80.9+100) * 4.184 * (82.2-27.5) = 54,264J
54,264/ 1000 = 54.264kJ
the answer is negative and 3 sigfigs
So, -54.3 kJ