At a certain temperature, the equilibrium constant, Kc, for this reaction is 53.3. H2(g) + I2...

At a certain temperature, the equilibrium constant, Kc, for this reaction is 53.3.

H2(g) + I2 (g) ---> 2 HI (g)

Kc= 53.3

At this temperature, 0.600 mol of H2 and 0.600 mol of I 2 were placed in a 1.00-L container to react. What concentration of HI is present at equilibrium?

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Expert Solution

For the given reaction -

H₂(g) + I₂ (g) ---> 2 HI (g)

I(M) 0.6 0.6 0

C -x -x +2x

Eq (0.6-x) (0.6-x) 2x

Therefore -

K_c = 53.3 = (2x)² / (0.6-x)²

By solving the above equation, the value of "x" is 0.470 M

Thus, at equilibrium, [HI] = 2(0.470 M) = 0.94 M

queen_honey_blossom answered 1 month ago

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