Question

Consider the chemical reaction below at a given temperature and at equilibrium: H2(g) +12(g) = 2HI(g) Kc = 53.3 If 0.800 mol of H2 and 0.800 mol of 12 are placed in a 1.00L container and allowed to react, what is the [HI] when the reaction reaches equilibrium? [HIN In the expression for K N- (H2] [12]' The equilibrium concentrations can be expressed as follows: NOTE: This is NOT asking for the concentrations you solve for this is literally asking you what you put in the ICE table in terms of x. (HI) - 2x (H2) = 0.8-x [12] = 0.8-X Now find the value of the equilibrium concentration of (HI) = [] Fullsc 1.26 M

Answer 1

According to the question ,

Kc = 53.3

Moles of H₂ = 0.800 moles

Moles of I₂ = 0.800 moles

Volume = 1 L

We know that Molarity = Number of Moles / Volume in L

Molarity of H₂ = 0.800 moles / 1 L = 0.80 M

Molarity of I₂ = 0.800 moles / 1 L = 0.800 M

H₂ + I₂ = 2 HI

So The expression for K = [HI]² / [H₂][ I₂]

So The Value of N = 2 .......( It is the Coefficient of HI in The Power format )

Now Make the ICE table

H₂ + I₂ = 2 HI

	H2 (M)	I2 (M)	2 HI (M)
Initial	0.80	0.80	0.00
Change	-x	-x	+ 2x
Equilibrium	0.80 - x	0.80 - x	+2x

So Value in the Terms of x ]

[H₂] = 0.80 - x

[ I₂] = 0.80 - x

[HI] = 2x

So K = [HI]² / [H₂][ I₂]

=> 53.3 = (2x)² / ( 0.80 -x ) ( 0.80-x )

=> x =  0.627 M

So equilibrium Concentration for HI = 2*x = 2*0.627 = 1.254 M

[HI] = 1.254 M

Thank you

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