Question

In: Chemistry

Find the value of the equilibrium constant Kc at 460°C for the reaction- ½ H2(g) +...

Find the value of the equilibrium constant Kc at 460°C for the reaction-
½ H2(g) + ½ I2(g)
HI(g)
given the following data: A 4.50-mol sample of HI is placed in a 1.00-L vessel at 460°C, and the reaction system is allowed to come to equilibrium. The HI partially decomposes, forming 0.343 mol H2 at equilibrium.
A) 0.0123 B) 0.0081 C) 0.0309 D) 11.1 E) 5.69

Expert Solution

Initial concentration of HI = number of moles / volume in L

= 4.50 mol / 1.00 L

= 4.50 M

HI(g) ½ H₂(g) + ½ I₂(g)

initial moles 4.50 0 0

change -a +a/2 +a/2

Equb moles 4.50-a a/2 a/2

Given Equilibrium moles of H₂ is = a/ 2 = 0.343 mol

a = 0.686 mol

So Equilibrium moles of HI = 4.50 - a = 4.50 - 0.686 = 3.814 moles

Equilibrium concentration of HI = number of moles / volume in L

= 3.814 mol / 1.0 L

= 3.814 M

Equilibrium concentration of H₂ & I₂ = 0.343 mol / 1.0L

= 0.343 M

Equilibrium constant , K = ([H₂(g)]^½ [I₂(g)]^½ ) / [HI]

= (0.343^½ x 0.343 ^½) / 3.814

= 0.0899

queen_honey_blossom answered 1 year ago

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