Answer the following questions on the Gibbs energy of oxidation reactions. — Physical Chemistry — Thermodynamics Chemistry Question
Gibbs energy of oxidation reaction
Answer the following questions on the Gibbs energy of oxidation reactions.
Fill (i) – (vi ) in the following sentences with suitable terms or chemical formulae.
When metal “M” is oxidized by one mole of oxygen gas to form an oxide MmOn (m, n: integers), the reaction can be expressed as (i). The standard Gibbs energy change of this reaction, ∆Go, can be expressed as (ii ) in terms of standard (iii ) change, ∆Ho, and standard (iv) change, ∆So , of this reaction at an absolute temperature T. On the other hand, when a pure metal M and a pure oxide MmOn are in equilibrium state at an absolute temperature T, the oxygen partial pressure pO2 can be derived as (v) using ∆Go and the gas constant R. The diagram where ∆Go values for various oxidation reactions are drawn as a function of absolute temperature is called an “Ellingham Diagram” (Figure 1). As can be seen in the figure, most relations are drawn by a straight line and the metals existing in the lower part tend to be (vi ) compared to those in the upper part.
[VISUAL]
Model Answer
(i) m/2n M + O2 = 2/n MmOn
(ii) ∆Ho – T∆So
(iii) enthalpy
(iv) entropy
(v) e(∆Go/RT)
(vi) oxidized
When both the reactants and the products are in the condensed state (solid or liquid) in Figure 1, the slope of each line in the diagram shows an almost identical value. The line is horizontal in the case of CO2 gas, and the slope shows a different sign with the same absolute value in the case of CO gas. Explain why.
Model Answer
In all the reactions except for the two C oxidations, 1 mole of the gas (oxygen), namely its entropy, is lost. This is why the slopes are almost identical. However, there are almost no changes in the entropy of the gas in the case of CO2 gas formation, making the line horizontal, and causing the increase in 1mole of the gas in the case of CO formation showing the different sign with the same slope.
Describe the chemical reaction when Cu2O is reduced by Al.
Model Answer
3 Cu2O + 2 Al = 3 Cu + Al2O3
Derive the heat of the reaction in the above question c) per mole of Al.
Model Answer
The heat generated (or absorbed) by the reaction per 1 mole of oxygen gas can be read from the difference in the values of the intercepts of the lines for Cu and Al. Hence, the value per 1 mole of Al can be obtained by multiplying 3/4. Then, the ∆Ho of the reaction can be read as –1130 – (–350) = –780 kJ per 1 mole of oxygen gas, and it will become –585 kJ per mole of Al. Heat of 585 kJ will be generated by the reaction. (exothermic reaction)
Show the points through which any lines of constant oxygen partial pressure pO2 and those with a constant value of the ratio of CO gas partial pressure to CO2 gas partial pressure pCO/pCO2 pass, respectively, in Figure 1.
Model Answer
Since the vertical axis shows RT ln pO2 value, any straight lines drawn through point “O” have the slope of R ln pO2. Hence, the value of pO2 is identical on such lines. On the other hand, the line of the reaction: 2 CO + O2 = 2 CO2 can be drawn by the two oxidation reactions of C mentioned in the question b), and the intercept of the derived line is assumed to be “C” through which the line of constant pCO/pCO2 value will go. They are plotted in the figure as “O” and “C”, respectively.
When solid FeO is reduced to Fe by flowing CO gas at 1000 K in equilibrium state, how much % of the CO gas will be consumed?
Model Answer
Two lines of 2 Fe + O2 = FeO and 2 CO + O2 = 2 CO2 happen to cross at 1000 K, which means that the value of the ratio of pCO/pCO2 is one when Fe and FeO coexist. Accordingly, the consumed fraction becomes 50 %.