Question

Potassium nitrate, KNO3, has a molar mass of 101.1 g/mol. In a constant-pressure calorimeter, 15.7 g of KNO3 is dissolved in 351 g of water at 23.00 °C.The temperature of the resulting solution decreases to 19.80 °C.

KNO3(s)+H2O(aq)>KOH(aq)KNO3(aq)

Assume the resulting solution has the same specific heat as water, 4.184 J/(g·°C), and that there is negligible heat loss to the surroundings. How much heat was released by the solution? and what is the enthalpy of the reaction?

Answer 1

KNO₃(s) + H₂O(aq) ------> KOH(aq) + KNO₃(aq)

Potassium nitrate, KNO3, has a molar mass of 101.1 g/mol.

mass of of KNO3 = 15.7 g

mass of water = 351 g of water

Tinitial = 23.00 °C

Tfinal = 19.80 °C.

Heat released by the solution Q = mass of solution x heat capacity of the liquid x change in Temp

Q = m x C x t

we can assume that heat capacity of calorimeter is same as that of solution and the solution has the same heat capacity of an equivalent amount of water (4.184 J/(g·°C)

Total mass of the solution = 15.7 + 351 = 366.7 g

t = |19.8 - 23.0 °C| = 3.2°C

Therefore,

Q = - [366.7 g x 4.184 J/(g·°C) x (3.2°C)]

= - 4909.67 J

Heat released by the solution , Q = 4909.67 J

Enthalpy of the reaction can be calculated using the formula,

Hrxn = Q/n

n = moles of KNO₃ dissolved

mass of of KNO₃ = 15.7 g

Molar mass of of KNO₃ = 101.1 g/mol.

Moles of KNO₃ dissolved = 15.7/101.1 = 0.1553 moles

Hrxn = Q/n = -4909.67 J/0.1553 moles = -31614.10 J/moles = -31.614 kJ/moles.