In a coffee-cup calorimeter, 1 mol KOH and 1 mol HBr initially at 27.5 o C (Cels
ID: 831329 • 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.
?H =
The equation for this reaction is:?Hrxn = -(C)(?T)(mass)
The entire contents of the calorimeter are changing temperature, so the mass includes the masses of the acid, base and water. The temperature of the contents is rising, so the reaction must be exothermic.
Explanation / Answer
Mass of KOH = moles x molar mass = 1 x 56.2 = 56.2 g
Mass of HBr = moles x molar mass = 1 x 80.9 = 80.9 g
Mass of water = 100 g
Mass of solution = 56.2 + 80.9 + 100 = 237.1 g
Delta Hrxn = -mass x specific heat x temperature change of solution
= -237.1 x 4.184 x (82.2 - 27.5)
= -5.43 x 10^4 J = -54.3 kJ