but i thgt that i read that under a pressure of abt 60000 atm and temp. over 1600 °C
graphite form changes into diamond.
So i thgt increase in pressure wuld favor backward reaction ///[1]
Q. densities of diamond and graphite are 3.51 and 2.25 g/cc respectively. increase in pressure on equilibrium C(diamond) <--> C(graphite)
(A) favours forward rkn
(b) " back "
(c) no effect
(d) increases rkn rate
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14 Answers
This is a highly endothermic equilibrium...the activation barrier for conversion of diamond to graphite is pretty high. So you know how to use Le Chatelier's principle on endothermic equilibriums..
favor's backward i guess ??? (b) ??
PRITISH here it's pressure not temperature for whic Le chatelier's principle for endothermic can be applied ??
Oh haan..lola. sorry.
But if there's no change in mols across the equilibrium, pressure change has no effect..diamond and graphite are structurally the same.
And you also think that a pressure of 60k atm is feasible in lab conditions..?
not laboratory conditions but is it anyways given that the reaction is performed in a laboratory
ANd by the way don't u find diamond in nature,
Howz that formed...[7]
That's formed in the earth, under pressure >> 60k atm.
Think about it. I'm not too sure by just applying lab grade pressure we can convert diamond to graphite. It's extremely tough.
arey yaar... y r we discussing abt earth ke andar kya hota hai??
pritish diamonds r manufactured in labs by graphite under high pre nd temp nd its true
Manmay and tanay both of you are talking about the reverse of this equilibrium...of course diamonds are manufactured that way. Who's even asking for that?? Its diamond to graphite, not graphite to diamond!
You're not understanding what I'm trying to say? You may manufacture diamond from graphite under suitable conditions in your lab/factory/your house but the conditions for diamond to graphite are much more stringent..the activation barrier is quite high and it requires even more temperature and pressure. Now if you want to make your house a furnace, that's upto you.
Now Manmay's ORIGINAL statement applies. we've gone too far arguing nothing. Pressure may favor backward equilibrium! but my doubt is that the overall mols remains same, its a structural change...why should pressure have to do anything at all favouring any side then(Le chatelier's principle)?
Well enough discussion has been made on this topic but none of you have given the answer with correct reason option B that is favour backward reaction is correct but can anyone give the correct explanation to it as per asked by pritish