anonymous
  • anonymous
Q. the graph relates In K vs 1/T for a reaction .the reaction must be :
Chemistry
chestercat
  • chestercat
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anonymous
  • anonymous
|dw:1354201232478:dw| A) exothermic B) endothermic
anonymous
  • anonymous
which option is correct?
anonymous
  • anonymous

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anonymous
  • anonymous
can u hlp me out .plzzz
ajprincess
  • ajprincess
i am really sorry. I have no idea
anonymous
  • anonymous
can u suggest any one on os to hlp me @ajprincess
ajprincess
  • ajprincess
I am nt sure. i rarely visit chemistry group. So have no idea. Am really sorry.
anonymous
  • anonymous
anonymous
  • anonymous
listen from the graph we find out the how K depends on temperature now we know that on increasing temp favourable for endothermic but here K i increasing on decreasing temp means it exothermic
anonymous
  • anonymous
might u read some law like on increasing temp exothermic show backward reaction while endothermic show forward reaction
anonymous
  • anonymous
temp is also increasing
anonymous
  • anonymous
no its not see 1/T
anonymous
  • anonymous
i'm nt understanding the graph ?
anonymous
  • anonymous
do u know wat is k?
anonymous
  • anonymous
euilibrium constant?
anonymous
  • anonymous
yes and will u give me the formula of kc
anonymous
  • anonymous
that is kc=concn of product/concn of react
anonymous
  • anonymous
ok now here in graph K is increasing on increasing the value of 1/T means temp i decreasing and on deacresing temp and K also increasing then it must be a exothermic process
anonymous
  • anonymous
in starting its temp. is 1.5 ,2,2.5 so its increasing
anonymous
  • anonymous
dear its not T u are provided 1/T
anonymous
  • anonymous
wat is 1/t
anonymous
  • anonymous
1/T means if the values are given are in increasing order den T is decreasing but if its value of T given in increassing order then T i increasing
anonymous
  • anonymous
decreasing then increasing
anonymous
  • anonymous
see whther they written T or 1/T or ask ur teach dey will explain u the difference abu T and 1/T
anonymous
  • anonymous
okay .
anonymous
  • anonymous
wait.
anonymous
  • anonymous
i hav understood it wait i'm posting 2nd question.

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