Question

calcium oxide reacts with water to form calcium hydroxide. when 11.1 g of cao completely reacts with 157.0 ml of h₂o in a coffee - cup calorimeter, the temperature increases from 18.0 °c to 37.4 °c.
what is the enthalpy of reaction (δhrxn) in kj/mol? assume no heat is lost to the surroundings, the density of water is 1.000 g/ml, and the specific heat of the solution is 4.184 j/g °c.

Explanation:

Step1: Calculate the mass of the solution

The mass of \(CaO\) is \(m_{CaO}=11.1\ g\). The volume of water \(V = 157.0\ mL\), and since density of water \(
ho=1.000\ g/mL\), the mass of water \(m_{H_2O}=
ho\times V=1.000\ g/mL\times157.0\ mL = 157.0\ g\).
The mass of the solution \(m=m_{CaO}+m_{H_2O}=11.1\ g + 157.0\ g=168.1\ g\)

Step2: Calculate the heat absorbed by the solution

Use the formula \(q = mc\Delta T\), where \(c = 4.184\ J/g^{\circ}C\), \(\Delta T=T_{final}-T_{initial}=37.4^{\circ}C - 18.0^{\circ}C=19.4^{\circ}C\), and \(m = 168.1\ g\)
\(q=(168.1\ g)\times(4.184\ J/g^{\circ}C)\times(19.4^{\circ}C)\)
\(q=168.1\times4.184\times19.4\ J\)
\(q = 168.1\times81.1696\ J\)
\(q=13644.6\ J\approx13.64\ kJ\)

Step3: Calculate the moles of \(CaO\)

The molar mass of \(CaO\) is \(M = 40.08\ g/mol+16.00\ g/mol = 56.08\ g/mol\)
The moles of \(CaO\), \(n=\frac{m}{M}=\frac{11.1\ g}{56.08\ g/mol}\approx0.198\ mol\)

Step4: Calculate the enthalpy of reaction

Since the reaction is exothermic (\(q_{system}=-q_{solution}\)), and \(\Delta H_{rxn}=\frac{q}{n}\)
\(\Delta H_{rxn}=\frac{- 13.64\ kJ}{0.198\ mol}\approx - 68.9\ kJ/mol\)

Answer:

\(-68.9\ kJ/mol\)