Given 2Al2O3 (s) --> 4Al(s) + 3O2 (g) (standard enthalpy change= 3351.4 kJ) a) What is the heat of formation of aluminum oxide
2Al2O3 (s) --> 4Al(s) + 3O2 (g) (standard enthalpy change= 3351.4 kJ)
a) What is the heat of formation of aluminum oxide?
How do I find heat of formation from standard enthalpy change? I know how to do it when I'm given enthalpy, but they aren't the same thing- correct?
Also, if this question gave me delta-H, I know that the heat of formation for 4Al = 0 and 3O2 = 0 because they are elementary species. This would leave me with 2Al2O3 = delta-H = x (whatever delta-H is) However, since I have 2 moles of 2Al2O3, would I have to divide delta-H by 2? Sorry if that was confusing.
The delta H for the reaction listed is just the reverse of the heat of formation. So add a negative sign to make it -3351 kJ for the reaction, then divide by 2 to make it 1/2*-3351 = 1676 kJ/mol.
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Given
2Al2O3 (s) --> 4Al(s) + 3O2 (g) (standard enthalpy change= 3351.4 kJ)
a) What is the heat of formation of aluminum oxide
Consider the reaction for the formation of aluminum oxide from aluminum and oxygen. 4Al(s)+3O2(g)⟶2Al2O3(s) ΔH1
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Consider the following reaction at 298 K. 4Al(s)+3O2(g)⟶2Al2O3(s) Δ?∘=−3351.4 kJ Calculate the following quantities. Refer to...
Consider the following reaction at 298 K. 4Al(s)+3O2(g)⟶2Al2O3(s) Δ?∘=−3351.4 kJ Calculate the following quantities. Refer to the standard entropy values as needed. Δ?sys= J/K Δ?surr= J/K Δ?univ= J/K Al Entropy = 28.3 O2 Entropy=205.2 Al2O3= 50.9
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