In a coffee-cup calorimeter, 1 mol KOH and 1 mol HBr initially at 26 o C (Celsiu

Question

In a coffee-cup calorimeter, 1 mol KOH and 1 mol HBr initially at 26 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 80.7 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 =

Explanation / Answer

we know that

mass = moles x molar mass

so

mass of KOH = 1 x 56.2 = 56.2

mass of HBr = 1 x 80.9 = 80.9

now

mass of solution = mass of KOH + mass oof HBr + mass of H20

so

mass of solution = 56.2 + 80.9 + 100

mass of solution = 237.1 g

now

heat = mass x specific heat x temp change

dH = m x s x dT

so

dH = 237.1 x 4.184 x (80.7 - 26)

dH = 54264 J

so

dH = 54.264 kJ

so

the change in enthalpy is 54.3 kJ

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects