Question

A calorimeter contains 24.0 mL of water at 11.0 ∘C . When 1.30 g of X (a substance with a molar mass of 66.0 g/mol ) is added, it dissolves via the reaction X(s)+H2O(l)→X(aq) and the temperature of the solution increases to 28.0 ∘C . Calculate the enthalpy change, ΔH, for this reaction per mole of X. Assume that the specific heat of the resulting solution is equal to that of water [4.18 J/(g⋅∘C)], that density of water is 1.00 g/mL, and that no heat is lost to the calorimeter itself, nor to the surroundings. Express the change in enthalpy in kilojoules per mole to three significant figures.

Answer 1

ΔH= 0.0924kJ/mol

Step-by-step explanation:

m= 1.3g, c= 4.18J/g°C, ∆t= 28-11= 17°C

Substitute into

ΔH=mc∆t = 1.3×4.18×17= 92.4J/mol = 0.0924kJ/mol

Answer 2

ΔH = -86.6 kJ/mol

Step-by-step explanation:

Step 1: Data given

Volume of water = 24.0 mL = 0.024 L

Temperature = 11.0 °C = 284 K

Mass of X = 1.30 grams

Molar mass X = 66.0 g/mol

The temperature of the solution increases to 28.0 ∘C

Specific heat of water = 4.18 J/g°C

Density = 1.00g/mL

Step 2: Calculate the heat transfer

Q = m*c*ΔT

⇒with Q = the heat transfer = TO BE DETERMINED

⇒with m = the mass = 24.0 mL * 1.00 g/mL = 24.0 grams

⇒with c= the specific heat of the water = 4.18 J/g°C

⇒with ΔT = the change of temperature = T2 - T1 = 28.0 °C - 11.0 °C = 17.0 °C

Q = 24.0 grams * 4.18 J/g°C * 17.0 °C

Q = 1705.44 J

Step 3: Calculate moles of X

Moles X = mass X / molar mass X

Moles X = 1.30 grams / 66.0 g/mol

Moles X = 0.0197 moles

Step 4: Calculate ΔH

Since the reaction is exothermic, ΔH is negative

ΔH = -Q/mol

ΔH = -1705.44 J / 0.0197 moles

ΔH = -86570.6 J/mol

ΔH = -86.6 kJ/mol