南邮材料化学答案-第四章习题答案

第四章材料化学热力学Chapter 4 Chemical Thermodynamics of Materials1. Cs熔体的标准吉布斯自由能（单位为J）与温度T（单位为K）的关系为=2100-6.95T, 求Cs的熔点。

Relationship between the standard Gibbs free ΔG om, Csenergy (in units of J) of Cs melt and temperature T (K) is shown as ΔG o m,=2100-6.95T, please determine the melting point of Cs.Cs解：根据热力学定律，当ΔG<0时，反应可以自发进行，因此ΔG＝0时对应的平衡反应（即Cs由固相转变为液相的反应）的温度即为Cs的熔点。

由ΔG o m, Cs=2100-6.95T＝0，可得T ＝302K，所以Cs的熔点为302K。

Answer: A ccording to the laws of thermodynamics, when ΔG <0, t he reaction can be spontaneous. Therefore, ΔG = 0 corresponding to the equilibrium reaction of Cs change from solid phase into liquid phase, the temperature corresponding to the melting point of Cs. From ΔG o m, Cs=2100-6.95T＝0, we can get T＝302K, so the melting point of Cs is 302K.2.通过埃灵罕姆图解释为何碳在高温下可以用作金属氧化物的还原剂。

According to the Ellingham diagram, please explain why the carbon can be used as a reductant of the metal oxides at high temperature.答：根据埃灵罕姆图，ΔG o-T曲线越在下方，氧化物的ΔG o负值越大，其稳定性就越高。

所以在给定的温度下，位于下方的ΔG o-T曲线所对应的元素可使位于上方曲线的金属氧化物还原。

由埃灵罕姆图还可知，CO生成线的斜率为负，随着温度升高，ΔG o越负，CO稳定性越高。

而金属氧化物的生成线的斜率都为正，随着温度升高，ΔG o越正，氧化物稳定性越低。

所以，碳在高温下可以用作金属氧化物的还原剂。

According to the Ellingham diagrams, the lower the site of ΔG o-T curve, the bigger the negative value of ΔG o, the more stable the metal oxide. So, at a given temperature, the element of the below ΔG o-T curve can reduce the metal oxide of the above ΔG o-T curve. We can see that the slope of CO formation line is negative, and with the increase of temperature, ΔG o becomes more negative, indicates that CO is more stable. However, the slopes of metal oxides formation line are all positive, with the increase of temperature, ΔG o becomes more positive, indicates that metal oxide is more unstable. Therefore, the carbon canbe used as a reductant of the metal oxides at high temperature.3.（a ）对任何反应，请说明ΔG 和ΔG o 为0的意义；（b ）对于“如果反应的ΔG o 是正值，则反应不会如反应式所写的那样向右进行”的说法，谈谈你的观点。

(a) for any reaction, please indicate the significance of ΔG 和ΔG o ; (b) For “if ΔG o of a reaction is positive, the reaction will not go along as the reactive written ”, talk about your opinion.答：（a ）对于任何反应，ΔG （吉布斯自由能变化）是描述一个化学反应的驱动力，即可据此判断反应发生的可能性。

根据热力学第二定律可知，“在任何自发变化过程中，自由能总是减少的”。

所以，ΔG 为0说明反应过程处于平衡状态。

而ΔG o 为标准吉布斯自由能变，即物质处于标准状态时（（指定温度（273.15 K ）和压力（101.325KPa ））的生成自由能。

（b ）已知ΔG o =－RTLn K ，若ΔG o 为正值，则Ln K 必需为负值，则K <1，表明反应进行的程度很小。

(a) for any reaction, ΔG (change of Gibbs free energy) is to describe driving force of a chemical reaction, you can judge the possibility of reaction. The second law of thermodynamics shows that the free energy is always reduced in any spontaneous reaction . Therefore, ΔG is 0 indicates that the reaction is in equilibrium. However, ΔG o stands for the standard Gibbs free energy change, that is to say, the material’s free energy of formation in the standard state (273.15 K and 101.325KPa).(b) As is known, ΔG o =－RTLn K , if ΔG o is positive, LnK must be negative, then K <1, indicating that the extent of reaction is very small. 4. 在埃灵罕姆图上大多数斜线的斜率为正，但是反应的斜率为0，反应的斜率为负，请解释原因。

In the Ellingham diagrams, slope of the most curves is positive, but the slope of the reaction C(s)+O 2(g)=CO 2(g) is zero, the slope of the reaction 2C(s)+O 2(g)=2CO(g) is negative, please explain the reason.答：对于C 氧化生成CO 2的反应，反应前后气体的分子数不变，因此熵变ΔS 为0，因为－ΔS 代表氧化物生成线的斜率，所以此时，－ΔS 为0，斜率为0。

对于C 氧化生成CO 的反应，反应以后气体分子数增加，即ΔS>0，－ΔS 为负，所以CO 生成线的斜率也为负。

1) For the reaction of C(s)+O 2(g)=CO 2(g), the number of gas molecules before and after 22C(s) +O (g)=CO (g)22C(s) +O (g)=2CO(g)the reaction is constant, so the entropy change ΔS is 0, because “－ΔS” stands for the slope of the formation line, in this case, ΔS is 0, so the slope is 0.2) For the reaction of 2C(s)+O2(g)=2CO(g), the number of gas molecules increase after the reaction, so the entropy change ΔS is >0, that is to say －ΔS<0, so the slope of the formation line is negative.5.在埃灵罕姆图中，哪些元素可以从二氧化硅中还原出硅？ In the Ellingham diagrams,which elements can reduce Si from SiO2?答：根据埃灵罕姆图，位于下方的ΔG o-T曲线所对应的元素可使位于上方曲线的金属氧化物还原。

TiO2、Al2O3、Li2O、CaO、MgO的生成线均位于SiO2生成线的下方，所以Ti、Al、Li、Ca、Mg这几种元素可以从二氧化硅中还原出硅。

According to the Ellingham diagrams, the corresponding element of the below ΔG o-T curve can reduce the corresponding metal oxide of the above curve. As is shown, the formation lines of TiO2、Al2O3、Li2O、CaO、MgO lie in the below of SiO2 formation line, so Ti、Al、Li、Ca、Mg can reduce Si from SiO2. Additional, above 1600℃, the formation line of CO locates below the formation line of SiO2, so C can reduce Si form SiO2 at temperature more than 1600℃.6.吉布斯相律通常为f=c-p+2，为什么在固体材料的研究中，相律一般可表达成f=c-p+1？Gibbs phase rule is usually expressed as f = c-p+2, why in the study of solid materials, the phase rule be expressed as f = c-p+1?答：在固体材料的研究中，压力对固相反应的影响很小，通常可以忽略，所以非成分的变量只有温度这一项，所以相律一般可表达成f=c-p+1。