A calorimeter contains 24.0 mL of water at 13.0 ∘C . When 2.00 g of X (a substance with a molar mass of 64.0 g/mol ) is added, it dissolves via the reaction
X(s)+H2O(l)→X(aq)
and the temperature of the solution increases to 26.5 ∘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 – Given, volume of water = 24 mL . temp, ti = 13.0 oC,
Mas of X = 2.00 g and its molar mass = 64.0 g/mole , tf = 26.5 oC
Density of water = 1.00 g/mL , specific heat = 4.18 J/goC
So the mass of water = 24 g
First we need to calculate the heat absorbed by water
We know formula for calculating the heat
q = m * C * ∆t
= 24 g * 4.18 J/goC * ( 26.5 -13.0 ) oC
= 1354.32 J
Now we know heat loss = heat gain
So, ∆H = -q
= - 1354.32 J
Now we need to calculate moles of X
Moles of X = 2.00 g / 64.0 g.mol-1
= 0.0313 moles
So the enthalpy change ΔH = - 1354.32 J / 0.0313 mole
= 43338.2 J/mol
= 43.3 kJ/mol
So enthalpy change, ΔH, for this reaction per mole of X is 43.3 kJ/mol
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