Question

0.50 g of mg reacts with excess oxygen in a bomb calorimeter. the overall heat capacity, including both the calorimeter and the water, is 3290 j/°c. the temperature changes from 21.5 °c to 25.2 °c. what is the enthalpy of the reaction? δh = ? kj/mol enter either a + or - sign and the magnitude. do not round until the end. use significant figures. use your units and watch your negatives.

Explanation:

Step1: Calculate heat absorbed by calorimeter

The heat absorbed by the calorimeter (\(q_{cal}\)) is given by \(q_{cal}=C\Delta T\), where \(C = 3290\ J/^\circ C\) and \(\Delta T=T_2 - T_1=25.2^\circ C - 21.5^\circ C = 3.7^\circ C\).
So, \(q_{cal}=3290\ J/^\circ C\times3.7^\circ C = 12173\ J\).

Step2: Determine heat of reaction (\(q_{rxn}\))

For a bomb calorimeter, the heat released by the reaction (\(q_{rxn}\)) is equal in magnitude but opposite in sign to the heat absorbed by the calorimeter. So, \(q_{rxn}=-q_{cal}=- 12173\ J=-12.173\ kJ\) (converted to kJ by dividing by 1000).

Step3: Calculate moles of Mg

The molar mass of Mg is \(24.305\ g/mol\). Moles of Mg (\(n\)) is given by \(n=\frac{mass}{molar\ mass}=\frac{0.50\ g}{24.305\ g/mol}\approx0.02057\ mol\).

Step4: Calculate enthalpy change (\(\Delta H\))

Enthalpy change is given by \(\Delta H=\frac{q_{rxn}}{n}\).
Substituting the values, \(\Delta H=\frac{- 12.173\ kJ}{0.02057\ mol}\approx - 591.7\ kJ/mol\).

Answer:

\(-592\) (rounded to three significant figures, the original calculation gives approximately - 591.7 which rounds to - 592)