Question

In a coffee-cup calorimeter, 1 mol NaOH and 1 mol HBr initially at 22.5 oC (Celsius) are mixed in 100g of water to yield the following reaction: NaOH + HBr → Na+(aq) + Br-(aq) + H2O(l) After mixing the temperature rises to 83 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 HBr is 80.9 g/mol.

Answer 1

Answer:

ΔH = -55.92 kJ

Explanation:

Step 1: Data given

1 mol NaOH and 1 mol HBr initially at 22.5 °C are mixed in 100g of water

After mixing the temperature rises to 83 °C

Specific heat of the solution = 4.184 J/g °C

Molar mass of NaOH = 40 G/mol

Molar mass of HBr = 80.9 g/mol

Step 2: The balanced equation

NaOH + HBr → Na+(aq) + Br-(aq) + H2O(l)

Step 3: mass of NaOH

Mass = moles * Molar mass

Mass NaOH = 1 * 40 g/mol

Mass NaOH = 40 grams

Step 4: Mass of HBr

Mass HBr = 1 mol * 80.9 g/mol

Mass HBr = 80.9 grams

Step 5: Calculate ΔH

ΔH = m*c*ΔT

ΔH= (100 + 40 + 80.9) * 4.184 * (83-22.5)

ΔH= 220.9 * 4.184 * 60.5

ΔH= 55916.86 J = 55.92 kJ

Since this is an exothermic reaction, the change in enthalpy is negative.

ΔH = -55.92 kJ