Homework #1 Binary System with a Peritectic 1800 The drawing shows the system KA

Question

Homework #1 Binary System with a Peritectic 1800 The drawing shows the system KAISi,O-SiO #A It includes the minerals leucite, K-spar, and three SiO2 polymorphs Melt 1600 Cristobalf (cristobalite, tridymite, + meltand quartz). Leucite + melt Note that the bottom axis is in wt%, not mole%. (This explains why K- spar does not plot half way between leucite and SiO2) 1 400 Tridymite p + melt 1200 K-spar +melt 1000 Leucite + K spar K-spar + tridymite K-spar+ quartzed 800 100 SiO2 80 KAISi2O6 20 leucite Wt% SiO2 KAS3Og K-spar Note the eutectic (point E) at about 1000 °C. Point E depicts the composition of the lowest temperature melt that can occur in this system. Congruent Melting. Incongruent Melting, and the Peritectic Leucite melts congruently at about 1680 °C; cristobalite melts congruently at about 1730 °C. This means that the melts produce are the same composition as the mineral that melts. (This diagram does not tell us anything about the melting temperature of tridymite or quartz, but they too melt congruently.) K-spar, in contrast with leucite and cristobalite, melts incongruently at 1150 °C to produce a liquid more siliceous than K-spar and the less siliceous compound, leucite. Point P, the composition of the melt when K-spar melts, is called a peritectic

Explanation / Answer

1. If we consider a melt composition F, it will start crystallize when it touches the liquidus curve at about 15200C. The first crystal will be leucite. On cooling, the melt will move along the liquidus curve with continuously changing the composition. When it will reach at point P (peritectic point), K-feldspar will start to form and at that point melt proportion decreases with forming leucite+ K-feldspar. The reaction is discontinuous here and K-felspar will form until there is no melt left.

2. The melt composition G will start forming crystal wwhen it touches the liquidus curve at about 16000C. The first forming crystal will be cristobalite. On cooling, it will move along the liquidus curve with forming tridymite. Finally it will reach at E(eutectic point), here crystallization of K-feldspar and tridymite will take place simultaneously but without any reaction relation between them.

3. Considering the rock composition at H point (50% KAlSi3O8 and 50% SiO2), the first melt will form when it touches the solidus curve and it will be composed of melt, K-feldspar and quartz. Now on heating to 18000C, it will melt and the melt composition will be changing through K-feldspar+leucite to leucite + melt composition. When it will reach at 18000 C, the melt will be only leucite rich composition.

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