A calorimeter contains 31.0 mL of water at 11.0 C . When 1.60 g of X (a substanc

Question

A calorimeter contains 31.0 mL of water at 11.0 C . When 1.60 g of X (a substance with a molar mass of 59.0 g/mol ) is added, it dissolves via the reaction

X(s)+H2O(l)X(aq)

and the temperature of the solution increases to 27.0 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.

Explanation / Answer

volume of water = 31 ml

mass of water = volume x density = 31 ml x 1g/ml = 31 g

substance mass = 1.6

solution mass = water mass + substance mass = 31 + 1.6 = 32.6 g

Initial temp = 11C , final Temp = 27 C

change in temp = 27-11 = 16 C

we have formula

Heat absorbed by solution = specific heat of solution x temp change x mass of solution

hence   Heat absorbed by solution = 4.184 J/gC x 16 C x 32.6 g = 2182.37 J

Heat absorbed by solution = Heat given by reaction

hence dH reaction = -2182.37 J

moles of X = mass/ molar mass of X = 1.6 /59 = 0.027

Hence when 0.027 moles added dH = -2182.37 J

for 1 mol of X we get dH = - ( 2182.37/0.027) = - 80828.5 J = -80.83 KJ

hence dH = - 80.83 KJ/mol

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