Answer to Question #350744 in General Chemistry for pluto

Question #350744
If 3.4 mol of HI was placed in a 2.0 L container and allowed to reach equilibrium, what would the equilibrium concentrations be for H2(g), I2(g) and HI(g) if the Ke = 49?	(5 marks)

H2(g) + I2(g) ⇆ 2HI(g)





1
Expert's answer
2022-06-15T23:12:03-0400


CM (HI) = n/V = 3.4mol / 2L = 1.7 mol/l

Ke= 49


2HI(g) → H2(g) + I2(g)

1.7-2x — x — x


"Ke= \\cfrac {[H_2][I_2]}{[HI]\u00b2}\\\\~\\\\\n49 = \\cfrac {x\u00b2}{(1.7-2x)\u00b2}\\\\~\\\\\n7 = \\cfrac{x} {(1.7-2x)}\\\\~\\\\\n11.9-14x=x\\\\\n15x=11.9\\\\x=0.793"


[HI] = 1.7 – 2*0.793 = 0.114 M

[H2] = 0.793 M

[I2] = 0.793 M








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