Question

The standard enthalpy of formation of mn2o3 is â962.3 kj/mol. how much heat energy is liberated when 7.9 grams of manganese are oxidized by oxygen gas to mn2o3 at standard state conditions? answer in units of kj.

Answer 1

Answer : -69.286 KJ heat energy is liberated.

Solution : Given,

Standard enthalpy of formation : When one mole of a substance is formed from its constituent elements.

$\Delta H_f=962.3KJ/mole$   (for 1 mole)

Mass of Mn = 7.9 g

Molar mass of Mn = 54.93 g/mole

First we have to calculate the moles of Mn.

$\text{ Moles of Mn}=\frac{\text{ Mass of Mn}}{\text{ Molar mass of Mn}}=\frac{7.9g}{54.93g/mole}=0.144moles$

The formation reaction of $Mn_2O_3$ is represented as,

$4Mn+3O_2\rightarrow 2Mn_2O_3$

For this reaction, the standard enthalpy of formation is,

$\Delta H_f=2\times 962.3KJ/mole=1924.6KJ/mole$

From the reaction, we conclude that

4 moles of Mn releases heat energy = -1924.6 KJ

0.144 moles of Mn releases heat energy = $\frac{-1924.6KJ}{4mole}\times 0.144mole=-69.286KJ$

Therefore, -69.286 KJ heat energy is liberated.