Question

A calorimeter contains 32.0 mLmL of water at 15.0 ∘C∘C . When 1.20 gg of XX (a substance with a molar mass of 54.0 g/molg/mol ) is added, it dissolves via the reaction X(s)+H2O(l)→X(aq)X(s)+H2O(l)→X(aq) and the temperature of the solution increases to 29.0 ∘C∘C . Calculate the enthalpy change, ΔHΔHDelta H, for this reaction per mole of XX. Assume that the specific heat of the resulting solution is equal to that of water [4.18 J/(g⋅∘C)J/(g⋅∘C)], that density of water is 1.00 g/mLg/mL, and that no heat is lost to the calorimeter itself, nor to the surroundings.

Answer 1

Answer:

The enthalpy change ΔH for this reaction per mole of X is -181.10 mol/kJ.

Explanation:

Volume of water in calorimeter = 32.0 mL

Mass of water in calorimeter = M

Density of the water = d = 1.00 g/mL

$M=1.00 g/mL\times 32.0 mL=32.0 g$

Mass of substance = 1.20 g

Mass of solution ,m= 1.20 g+ 32.0 g = 33.20 g

Specific heat of solution = c = 4.18 J/g°C

Change in temperature of the calorimeter = ΔT = 29.0°C

Heat absorbed by the solution = Q

$Q=m\times c\times \Delta T$

$Q=33.20 g\times 4.18J/g^oC\times 29.0^oC=4,024.504 J$

Moles of substance  X added to water ,n= $\frac{1.20 g}{54.0 g/mol}=0.02222 mol$

Heat released when 1.20 grams of X was dissloved = Q'= -Q = -4,024.504 J

Q' = -4,024.504 J = -4.024504 kJ ≈ 4.0245 kJ

1 J = 0.001 kJ

Enthalpy change for this reaction per mole of X:

$\Delta H=\frac{Q'}{n}=\frac{-4.0245 kJ}{0.02222 mol}=-181.10 mol/kJ$