Calculate the enthalpy of formation of maganese (IV) oxide based on the following information.
4Al(s) + 3MnO2(s) --> 3Mn(s) + 2Al2O3(s) Delta H = -1790 kJ
2Al(s) + 3/2O2 (g) ---> Al2O3 (s) Delta Hf = -1676 kJ
we need twice the second reaction to make 2Al2O3, then the first reaction reversed to make three moles of MnO2
Hr=2*(-1676kJ)+1790kJ
but we make 3 moles MnO2
hf=Hr above/3 kJ/mole
To calculate the enthalpy of formation of manganese (IV) oxide (MnO2), we can use the given thermochemical equation and the enthalpy of formation of aluminum oxide (Al2O3).
The enthalpy change for the reaction:
4Al(s) + 3MnO2(s) → 3Mn(s) + 2Al2O3(s) ΔH = -1790 kJ
can be written as the sum of the enthalpy changes for the individual reactions:
4Al(s) + 3/2 O2(g) → 2Al2O3(s) ΔH₁
and
3/2 O2(g) → MnO2(s) ΔHf(MnO₂)
We are given the ΔH₁ for the reaction:
2Al(s) + 3/2 O2(g) → Al2O3(s) ΔH₁ = -1676 kJ
We need to find the enthalpy of formation of MnO2 (ΔHf(MnO₂)), but we know the enthalpy change for the reaction:
3/2 O2(g) → MnO2(s)
To find ΔHf(MnO₂), we can reverse the sign and adjust the stoichiometry by multiplying the equation by 2/3:
MnO2(s) → 3/2 O2(g) ΔHf(MnO₂) = - (2/3) ΔH₂
Now, we can substitute the values into the equation:
-1790 kJ = ΔH₁ + (2/3) ΔH₂
Solving for ΔH₂ (enthalpy of formation of MnO2), we get:
ΔH₂ = (-1790 kJ - ΔH₁) / (2/3)
Substituting the values:
ΔH₂ = (-1790 kJ - (-1676 kJ)) / (2/3)
ΔH₂ = (-1790 kJ + 1676 kJ) / (2/3)
ΔH₂ = -114 kJ / (2/3)
ΔH₂ = -171 kJ
Therefore, the enthalpy of formation of manganese (IV) oxide (MnO2) is -171 kJ/mol.
To calculate the enthalpy of formation of manganese(IV) oxide (MnO2), we need to use the given chemical equation and the enthalpy of formation of aluminum oxide (Al2O3).
The given equation is:
4Al(s) + 3MnO2(s) --> 3Mn(s) + 2Al2O3(s) ΔH = -1790 kJ
The equation also states the enthalpy change (ΔH) for this reaction, which is -1790 kJ.
We are also given the enthalpy of formation of aluminum oxide (Al2O3):
2Al(s) + 3/2O2(g) ---> Al2O3(s) ΔHf = -1676 kJ
To calculate the enthalpy of formation of MnO2 (ΔHf(MnO2)), we can use the concept of Hess's Law. Hess's Law states that if we can express a target reaction as a combination of other known reactions, then the enthalpy change of the target reaction can be calculated by algebraically manipulating the known reactions.
First, we need to manipulate the first equation so that it has the same reaction as the given enthalpy of formation equation for Al2O3. To do this, we multiply the first equation by 2, and reverse the equation for the formation of Al2O3. This gives us:
2Al2O3(s) --> 4Al(s) + 6MnO2(s) ΔH = 3580 kJ (multiplied the first equation by 2 and reversed the equation)
Now, we can sum up the two equations to cancel out the common species (4Al(s) and 2Al2O3(s)) and obtain the target reaction:
2Al2O3(s) + 6MnO2(s) --> 3Mn(s) + 2Al2O3(s) ΔH = -1790 kJ
2Al2O3(s) --> 4Al(s) + 6MnO2(s) ΔH = 3580 kJ
By adding these two equations, we get:
2Al2O3(s) + 6MnO2(s) --> 3Mn(s) + 2Al2O3(s) + 2Al2O3(s) ΔH = -1790 kJ + 3580 kJ
2Al2O3(s) + 6MnO2(s) --> 3Mn(s) + 4Al(s) + 4Al2O3(s) ΔH = 1790 kJ
Now, we see that the target reaction is:
6MnO2(s) --> 3Mn(s) + 4Al(s) + 4Al2O3(s)
We want to calculate the enthalpy of formation of MnO2 (ΔHf(MnO2)). To do this, we can divide the target reaction by 6 to isolate the formation of MnO2:
MnO2(s) --> Mn(s) + (4/6)Al(s) + (4/6)Al2O3(s) ΔH = 1790 kJ / 6
MnO2(s) --> Mn(s) + (2/3)Al(s) + (2/3)Al2O3(s) ΔH = -298.3 kJ
So, the enthalpy of formation of manganese(IV) oxide (MnO2) is approximately -298.3 kJ.
Remember to double-check this calculation and consider any rounding errors that may have occurred.