Question

A 54.7 g sample of arsenic tribromide was heated until the compound melted. The molten compound was then poured into a calorimeter containing 300.0 g water at 22.50 °C. When the last bit of the compound had solidified, the temperature of the water was 24.13 °C. This is called the molar heat of fusion of AsBr3. Calculate ΔH (per mole of AsBr3)

Answer 1

ΔH = - 11758.6 J/mol = - 11.76 kJ/mol

Step-by-step explanation:

Step 1: Data given

Mass of arsenic tribromide = 54.7 grams

MAss of water = 300.0 grams

Temperature of water = 22.50 °C

Final temperature of water = 24.13 °C

Step 2: Calculate ΔH

Q= m*c*ΔT

⇒with m =the mass of water = 300.0 grams

⇒with c= the specific heat of water = 4.184 J/g°C

⇒with ΔT = the change in temperature of water = T2 - T2 = 24.13 - 22.50 = 1.63 °C

Q = 300 * 4.184 * 1.63

Q = 2046 J

Step 3: Calculate moles AsBr3

Moles AsBr3 = mass / molar mass

Moles AsBr3 = 54.7 grams / 314.634 g/mol

Moles AsBr3 = 0.174 moles

Step 4: Calculate ΔH

ΔH = -Q/moles

ΔH = -2046 J / 0.174 moles

ΔH = - 11758.6 J/mol = - 11.76 kJ/mol