24. The equilibrium constant for the reaction represented below is 50 at 448°0 H2(g) +12(g) 2...

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science chemistry

Answer 1

Answer #1

	H₂	I₂	2HI
initial number of moles	1	1	----
Number of moles reacted	x	x	2x
Number of moles remaining at equilibrium	1-x	1-x	2x
Equilibrium concentration	(1-x)/V	(1-x)/V	(2x)/V

Given K_c=50,

According to Law of mass action,

K_c = [HI]²/[H₂][I₂]

50 =(2x)²/(1-x)²

X= 0.78

Therefore [HI] =(2x0.78)/0.5

=3.12 mole L^-1

Answer 2

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