Question

potassium nitrate, kno₃, has a molar mass of 101.1 g/mol. in a constant - pressure calorimeter, 31.6 g of kno₃ is dissolved in 337 g of water at 23.00 °c.
kno₃(s) xrightarrow{h₂o} k⁺(aq) + no₃⁻(aq)
the temperature of the resulting solution decreases to 21.50 °c. assume that 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?
q_{soln}=
what is the enthalpy of the reaction?
δh_{rxn}=

Explanation:

Step1: Calculate the mass of the solution

The mass of the solution \(m = m_{KNO_3}+m_{water}\). Given \(m_{KNO_3}=31.6\ g\) and \(m_{water} = 337\ g\), so \(m=(31.6 + 337)\ g=368.6\ g\)

Step2: Calculate the heat released by the solution

Use the formula \(q = mc\Delta T\), where \(c = 4.184\ J/(g\cdot^{\circ}C)\), \(\Delta T=T_{final}-T_{initial}\). Here \(T_{initial}=23.00^{\circ}C\), \(T_{final}=21.50^{\circ}C\), so \(\Delta T=(21.50 - 23.00)^{\circ}C=- 1.50^{\circ}C\)

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The heat released by the solution is \(|q_{soln}| = 2.31\ kJ\)

Step3: Calculate the number of moles of \(KNO_3\)

Use the formula \(n=\frac{m}{M}\), where \(m = 31.6\ g\) and \(M = 101.1\ g/mol\)

$$ n=\frac{31.6\ g}{101.1\ g/mol}=0.3126\ mol $$

Step4: Calculate the enthalpy of the reaction

Use the formula \(\Delta H=\frac{q}{n}\). Since \(q = 2.31\ kJ\) (heat absorbed by the reaction, because the solution releases heat) and \(n = 0.3126\ mol\)

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Answer:

\(q_{soln}=2.31\ kJ\)

\(\Delta H_{rxn}=7.39\ kJ/mol\)