A calorimeter contains 35.0 mL of water at 12.0 C . When 2.00 g of X (a substanc

Question

A calorimeter contains 35.0 mL of water at 12.0 C . When 2.00 g of X (a substance with a molar mass of 72.0 g/mol ) is added, it dissolves via the reaction and the temperature of the solution increases to 29.0 C .

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

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

Solution :-

Volume of water = 35 ml = 35 g    (d=1 g/ml)

Initial temperature T1 = 12.0 c

Final temperature T2 = 29.0 C

Specific heat of solution = 4.18 J per g C

Lets first calculate the amount of heat produced by the 2.00 g substance X

q= m*s*delta t

q= 35 g * 4.18 J per gC * (29.0 C – 12.0 C)

q= 2487 .1 J

now lets calculate the amount of energy given by the 1 mole of the substance X

72.0 g per mol * (2487.1 j) / 2.00 g = 89536 J

Since reaction is exothermic therefore it has negative sign

Lets convert it to kJ

-89536 J * 1 kJ / 1000 J = -89.5 kJ per mol

The sign is negative because the reaction is exothermic so it looses heat

Therefore the enthalpy change per mole of X is - 89.5 kJ per mol

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