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In: Chemistry

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?

Solutions

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 -

                                      Kc = 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 3 months ago
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