A calorimeter contains 18.0 mL of water at 14.0 C . When 2.40 g of X (a substanc

Question

A calorimeter contains 18.0 mL of water at 14.0 C . When 2.40 g of X (a substance with a molar mass of 67.0 g/mol ) is added, it dissolves via the reaction X(s)+H2O(l)X(aq) and the temperature of the solution increases to 29.5 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.

Explanation / Answer

Q = m Cp dT of the calorimeter + substance
Q = [18.0mLx1.00g/1mL + 2.50g] x 4.18 J/g°C x (29.5°C - 14.0°C) = 1328.195 J

and that is on a per 2.40g basis.. so we need to convert to moles
(1328.195J / 2.40g) x (67.0g / mole) x (1kJ / 1000J) = 37.1 kJ/mole

the enthalpy change, H = -37.1 kJ/mole

remember that in this case H is negative because H is always the opposite of the q you calculated because the q you calculated is the q for the surround and not system. The H that you want to find is q system that is why the H have an opposite sign of the q.

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