Physical Chemistry Please help, not quite sure how to work through this problem.
ID: 1034881 • Letter: P
Question
Physical Chemistry
Please help, not quite sure how to work through this problem.
The following cooling curves have been found for the system antimony-cadmium Cdwr. % First break in curve, °C Constant temperature, °C 630 410 410 410 410 439 295295 0 20 37.5 47.550 58 7093 100 550 461 -419 400-- 321 Construct a phase diagram, assuming that no breaks other than these actually occur in any cooling curve. Label the diagram completely and give the formula of any compound formed. How many degrees of freedom are there for each area and at each eutectic point?Explanation / Answer
Let us draw phase diagram of Sb-Cd system using the above given informations.
*take percentage composition of Cd from 0% to 100 % as X-axis and temperature on Y-axis.
* mark temperature points 630,550,410,461,419,439,400,295,321 with respective of corresponding compositions.( Datas given above)
* Complete the graph joining the points.
* From the graph we can see, 630°c at 0% Cd is the melting point of pure Sb and 321°c at 100% Cd is the melting point of pure Cd.
The four points 550,461,419 and 400 lies on the curves. There are two depression points at 410 °c and 295 °c at 47.5% and 93 % respectively and a maximum point at 439°c of 58%..
*label the points A(630),B(321),C(410),D(439),E(295).
*curves AC, BE, and CDE are the respective freezing point curves of Sb, Cd, and compount formed by Sb-Cd.. These curves represent two component and two phase equilibria.
Therefore degrees of freedom F=C-P+1 = 2-2+1 = 1.
That is monovariant.
*area above three curves represents the existence of homogeneous liquid mixture of Sb and Cd.
F=C-P+1 =2-1+1 =2.
That is bivariant
* areas below the curves represence the presence of one solid phase and its liquid phase.
F=C-P+1 =2-2+1 =1
That is mono variant
*the two depression points are the eutectic points of the system. Representing 3 phase equilibria.
F=2-3+1 =0.
That is the system is invarient.
*Sb and Cd forms stable compound at the maximum point D. The formula of the compound can be find out
The compound forms at 58% Wt of Cd.
That is 100g of compound contains 58g of Cd and 42g Sb.
Molar mass of Cd and Sb are 112.411 g/mol and 121.757 g/mol.
Thus the mole ratio of Cd and Sb is given by
58/112.411 : 42/121.757
That is 0.51596 of Cd and 0.34495 of Sb.
Dividing by small number.,gives
1.5 of Cd and 1 of Sb.
Hence the molecular formula of the compound formed by Cd and Sb is Cd3Sb2.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.