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?

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:

answer: the rate of diffusion increases with increase in temperature.

increase in temperature, increases the molecular/particles speed (kinetic energy. diffusion means movement of particles through the material.

generally, the heavier the particle the more it interacts with the environment and more the temperature matters. so higher temperature results in higher diffusion.

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