Tianjin University Practice Test (1), Question 19
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19. According to phase rule analysis of the reaction system in which CaCO3 decomposes into CaO and CO2, under a CO2 atmosphere at a specified pressure, the degree of freedom of the equilibrium system is ( ). A. F=0 B. F=1 C. F=2 D. F=3 How should this question be approached? Thank you
This post was last edited by liujia1987 on 2015-8-5 09:01. There are 3 substances in this system, so S=3. One reaction is the decomposition of CaCO3 into CaO and CO2; thus R=1. The ratio of CaO to CO2 is 1:1, so R′=1 as well. Therefore, C=S-R-R′=1. There are solids and gases in the system, so P=2. Since the pressure remains constant, F=C-P+1=1-2+1=0. I’m not sure if it’s correct; please give me some advice.
There are 3 substances in this system: S=3, R=1. The ratio of CaO to CO2 is 1:1, so R′=1. Therefore, C=S-R-R′=1. There are solids and gases in the system, so P=2. Then F=C-P+1 (note that it refers only to pressure), which equals 1-2+1=0. Hence, the answer is (A). Note the meaning of n in Gibbs’ phase rule, f=C-P+n!
r1 is equal to zero, p waits for 3. two
Thank you for the advice; I ignored keeping the pressure constant