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Use the following data to calculate the standard heat (enthalpy) of formation, DH°f , of manganese(IV)...

Use the following data to calculate the standard heat (enthalpy) of formation, DH°f , of manganese(IV) oxide, MnO2 (s).

2MnO2(s) ® 2MnO(s) + O2(g) DH = 264 kJ

MnO2(s) + Mn(s) ® 2MnO(s) DH = -240 kJ

A. -24 kJ

B. 24 kJ

C. -372 kJ

D. 504 kJ

E. -504 kJ

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Answer #1

Balanced equation for formation of MnO2(s) in its standard state is

Mn(s) + O2(g) -----------> MnO2(s) --------Eq(1)

Given reactions are

2MnO2(s) -----------> 2MnO(s) + O2(g)    DH = + 264 kJ   -----------Eq (2)

MnO2(s) + Mn(s) -------> 2MnO(s)           DH = -240 kJ --------- Eq(3)

We have to rearrange Eqs (2) and (3) to get Eq(1).

Since O2(g) is reactant in Eq(1), write the reverse reaction of Eq(2)

2MnO(s) + O2(g) -------> 2MnO2(s)    DH = - 264 kJ

Since Mn(s) is reactant in Eq(1), rewrite the same reaction in Eq(3)

MnO2(s) + Mn(s) -------> 2MnO(s)           DH = -240 kJ

---------------------------------------------------------------------------------

2MnO(s) + O2(g) -------> MnO2(s) + MnO2(s)    DH = - 264 kJ

MnO2(s) + Mn(s) -------> 2MnO(s)                      DH = -240 kJ

--------------------------------------------------------------------------------

Cancel the similar terms on both sides and sum the all DH.

Then, the resultant reaction is

Mn(s) + O2(g) -----------> MnO2(s)        DH= - 264 kJ + (-240 kJ) = -504 kJ

Therefore,

standard heat (enthalpy) of formation, DH°f , of manganese(IV) oxide, MnO2 (s) = -504 kJ

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