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  • The equilibrium constant for the following reaction is 1.6 ×10^5 at 1024K H_2(g) + Br_2(g) 2HBr(g) Find the equilibrium pressure of all gases if 10.0

7.27     The equilibrium constant for the following reaction is 1.6 ×105 at 1024K

               H_{2}(g)+Br_{2}(g)\rightleftharpoons 2HBr(g)

Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K.

Answers (1)

best_answer

Given that,
K_p for the reaction = 1.6\times 10^{5}

K'_p = 1/K_p = \frac{1}{1.6\times 10^{5}} = 6.25 \times 10^{-6}

Let the pressure of both H_2 and Br_2  at equilibrium be p.

                        2HBr\rightleftharpoons H_2(g)+Br_2(g)
initial conc.           10                   0              0
at eq                     10-2p             p                p

Now,
  K_p' = \frac{p^{Br_2}.p^{H_2}}{p^2_{HBr}}
6.25\times 10^{-6}=\frac{p^2}{(10-2p)^2}
5\times 10^{-3}=\frac{p}{(5-p)}
By solving the above equation we get, 

p = 0.00248 bar

Hence the pressure of  H_2 and Br_2  is 0.00248 bar and pressure of HBr is 0.00496 bar

Posted by

manish

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