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# unit 2 watch

1. Hydrogen and iodine gases were mixed at 300°C and allowed to reach equilibrium.
H2(g) + I2(g) = 2HI(g) ǻH = –10 kJ mol–1
Colorless Purple Colourless
(a) What would you see if the equilibrium mixture was cooled to 250°C and
equilibrium allowed to re-establish?
(1) A The mixture goes a darker purple.
B The colour gets lighter.
C The mixture goes colourless.
D No visible change
2. (Original post by saif_ali)
Hydrogen and iodine gases were mixed at 300°C and allowed to reach equilibrium.
H2(g) + I2(g) = 2HI(g) ǻH = –10 kJ mol–1
Colorless Purple Colourless
(a) What would you see if the equilibrium mixture was cooled to 250°C and
equilibrium allowed to re-establish?
(1) A The mixture goes a darker purple.
B The colour gets lighter.
C The mixture goes colourless.
D No visible change
well the forward reaction is exothermic ( enthalpy change is negative ). So the reverse reaction is endothermic right?
If we decrease the temp (from 300 to 250 degrees celsius) then the system will oppose the increase in temperature by favouring the exothermic reaction (as it gives out heat). This means the system will decrease the temperature. so equilibirum will shift to the right as the forward reaction(exothermic reaction) is favoured.
Therefore what will happen? there will be a higher partial pressure of hydrogen iodide and there will be lesser iodine (as more of it was used in the forward reaction). HI is colourless whereas iodine vapour is purple

Conclusion = the colour gets lighter so B is the answer.

Looking at the other options...

A is wrong because if colour gets darker then that implies that there is a higher partial pressure of iodine ( so more iodine ) which clearly is wrong.

C is wrong because it implies there is no iodine left at all. Although its true that more iodine and hydrogen was used up when the forward reaction was favoured, there will still be some iodine vapour left. There will be lesser iodine vapour, but that's not the same thing as saying that there is no iodine left.

D is clearly wrong as it implies no visible reaction occurred.

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