Question

Potassium nitrate, KNO3, has a molar mass of 101.1 g/mol. In a constant-pressure calorimeter, 32.3 g of KNO3 is dissolved in 243 g of water at 23.00 °C.KNO3(s)+H2O(aq) ---> KOH(aq)+HNO3(aq)The temperature of the resulting solution decreases to 17.90 °C. 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.1. How much heat was released by the solution?2. What is the enthalpy of the reaction?

Answer 1

Answer:

1. Qsol = -5.87 kJ

2. ΔHrn = 18.4 kJ/mol

Explanation:

According to the law of conservation of energy, the sum of the heat absorbed by the reaction and the heat released by the solution is zero.

Qrn + Qsol = 0

Qrn = -Qsol

We can calculate the heat released by the solution using the following expression.

Qsol = c . m . ΔT

where,

c: specific heat capacity of the solution

m: mass of the solution

ΔT: change in the temperature

Qsol = (4.184 J/g.°C) . (243g + 32.3g) . (17.90°C-23.00°C) = -5.87 × 10³ J = -5.87 kJ

The heat absorbed by the reaction is:

Qrn = -Qsol = 5.87 kJ

In the balanced equation, we have 1 mole of KNO₃. Given we are in a constant-pressure calorimeter, the enthalpy of reaction (per mole of KNO₃) is:

$\Delta Hrn = \frac{5.87kJ}{32.3g}.\frac{101.1g}{mol} =18.4 kJ/mol$