A generic solid, X, has a molar mass of 67.9 g/mol. In a constant-pressure calorimeter, 16.8 g of X is dissolved in 269 g of water at 23.00 °C.
X(s) ------ X(aq)
The temperature of the resulting solution rises to 26.40 °C. Assume the solution has the same specific heat as water, 4.184 J/(g·°C), and that there\'s negligible heat loss to the surroundings.
1. How much heat was absorbed by the solution?
2. What is the enthalpy of the reaction?
1) Heat absorbed by the solution
Heat absorbed = m Cp (T2 - T1)
m = Mass = 269 + 16.8 = 285.8 g
Cp = 4.184
T2 = 26.4
T1 = 23
=> Heat absorbed = 285.8 x 4.184 x (26.4 - 23) = 4065.7 J
Moles of X = 16.8 / 67.9 = 0.2474 moles
Enthalpy of a reaction is defined as Heat evolved per mole of the reaction
For 0.2474 moles of X, heat evolved = 4065.7 J
=> For 1 mole of X, Heat evolved = 4065.7 / 0.2474 = 16432.2 J
=> Enthalpy of reaction = - 16432.2 J / mol or - 16.43 kJ / mol (-ve since heat is being evolved)
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