(activity in terms of molality) (van\'t Hoff equation) (integrated form of van\'
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(activity in terms of molality) (van't Hoff equation) (integrated form of van't Hoff T T2 equation) (logarithmic form of K) 5.3. Which system in each pair best represents equilibrium spe- cies under standard conditions of temperature and pressure? ys chYou may have to apply some basic knowledge of chemistry (a)Rb & H20 or Rb& OH & H2 (b)Na & Cl2 or NaCI (crystal) (c) HCI & H,0 or H (aq) &CI; (aq) ? (d)C (diamond) or C (graphite) Unless otherwise noted, all art on this page is © Cengage Learning 2014.Explanation / Answer
Answer:- A) In first Rb+, OH- and H2 represents the equilibrium because Rb metal is reactive. It comes in the first group and sixth period it reacts very violently with water and forms hydroxide and eliminates hydrogen and this happens because its electronic configuration is [Kr]366S1 so after losing one electron it will attain the noble gas configuration, which is very stable and it is in period 6 so the outer most electron is loosely bound compared to metal in above period that's why its reaction with water is even faster than the reaction of potassium with water.so when Rb will come in contact with water it will lose one elecron easily and will reduce water to OH- and H2 and itself will get oxidise to Rb+ hence Rb+, OH- and H2 will be the species in equilibrium.
B) IN this NaCl crystal represents the equilibrium species because Na and Cl2 react very violently and gives NaCl. Reason:- the reason behind this is the very negative lattice enthalpy (-787.3 kJ/mol) and electron affinity of Cl (-348.8 kJ/mol) so -1136.1 kJ/mol energy will liberate which will overcome all the reaction barrier.
C) H+(aq) and Cl-(aq) represents the equilibrium species because of very high hydration enthalpy of H+ and Cl- so the moment HCl passed through water it splits into the hydrated ion H+(aq) and Cl-(aq).
D) Between graphite and diamond, graphite is thermodynamically more stable because, In graphite, carbon is sp2 hybridized and possess free electron which is delocalized through it makes more conjugation whereas in diamond, carbon is sp3 hybridized and there is no free electrons in diamond so because of delocalization of electron graphite is thermodynamically more stable so graphite should be the species present at equilibrium but the equilibrium for the species does not occur very fast means if we have diamond it will not convert to graphite and same as graphite will not convert to diamond, the equilibrium time is very long for that reacton to happen.
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