A calorimeter contains 24.0 mL of water at 13.0 C . When 2.00 g of X (a substanc

Question

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, H, 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/(gC)], 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.

Explanation / Answer

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 H = - 1354.32 J / 0.0313 mole

                                       = 43338.2 J/mol

                                      = 43.3 kJ/mol

So enthalpy change, H, for this reaction per mole of X is 43.3 kJ/mol

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