Question

The equilibrium constant for the following reaction is $ 1.6 \times {10^5} $ at 1024K.
  $ {H_{2(g)}} + B{r_{2(g)}} \rightleftharpoons 2HB{r_{(g)}} $
Find the equilibrium pressure of all gases if $ 10.0bar $ of HBr is introduced into a sealed container at 1024K.

Answer 1

Hint :The equilibrium constant $ {K_p} $ gives us a relationship between the partial pressures of reactants and products and is calculated using the partial pressures of the reaction equation. It is Unit less quantity.

Complete Step By Step Answer:
We are given the $ {K_p} $ value for the reaction $ {H_{2(g)}} + B{r_{2(g)}} \rightleftharpoons 2HB{r_{(g)}} $ which is $ 1.6 \times {10^5} $ . The $ {K_p} $ values for the inverse reaction $ 2HBr \rightleftharpoons {H_{2(g)}} + B{r_{2(g)}} $ $ K{'_p} $ will be the inverse of $ {K_p} $
Therefore we can say that $ K{'_p} = \dfrac{1}{{{K_p}}} = \dfrac{1}{{1.6 \times {{10}^5}}} = 6.25 \times {10^{ - 6}} $
We are given that the initial pressure is $ 10.0bar $ . Let ‘p’ be the change in pressure at final/equilibrium change. The reaction can be give as:
                              $ 2HBr \rightleftharpoons {H_{2(g)}} + B{r_{2(g)}} $
Initial 10 - -
Final $ 10 - 2p $ $ p $ $ p $
The value of $ K{'_p} $ can be given as: $ K{'_p} = \dfrac{{{p_{{H_2}}} \times {p_{B{r_2}}}}}{{p_{HBr}^2}} $
  $ K{'_p} = \dfrac{{p \times p}}{{{{(10 - 2p)}^2}}} = 6.25 \times {10^{ - 6}} $
  $ \dfrac{p}{{10 - p}} = 2.5 \times {10^{ - 3}} $
  $ p = 2.5 \times {10^{ - 2}} - (5.0 \times {10^{ - 3}})p $
  $ p + (5.0 \times {10^{ - 3}})p = 2.5 \times {10^{ - 2}} $
  $ p = 2.49 \times {10^{ - 2}}bar \approx 2.5 \times {10^{ - 2}}bar $
The partial pressures of Hydrogen and Bromine gas is equal to p, hence the partial pressures of
  $ {p_{{H_2}}} = {p_{B{r_2}}} = p = 2.5 \times {10^{ - 2}}bar $
  $ {p_{HBr}} = 10 - 2p = 10 - 2(2.5 \times {10^{ - 2}}) \approx 10bar $ (Since the value of 2p is very small we can neglect it)
Therefore the partial pressures of all the gases are found.
Remember that in an arithmetic equation if the value (comparatively larger) is added or subtracted from a value $ \leqslant 0.05 $ that value can be neglected, since it will cause negligible change in the overall answer.

Note :
Always remember that the equilibrium constant of an inverse reaction (inverse to the one given in the question) will be the reciprocal of the original equilibrium constant. This is applicable for both $ {K_p}\& {K_c} $ . For example consider the following reaction:
  $ A + B \rightleftharpoons AB $ $ {K_p} $ $ {K_c} $
The equilibrium constant for the reverse reaction, $ AB \rightleftharpoons A + B $ denoted by $ K{'_c} $ can be given as:
   $ K{'_c} = \dfrac{1}{{{K_c}}} $
The terms $ {K_p}\& {K_c} $ are related to each other by a simple equation, $ {K_p} = {K_c}{(RT)^{\Delta n}} $
Where, $ \Delta n $ is the difference in the moles of product and reactant. T is the temperature and R is the gas constant.