question_answer 1)

The equilibrium constant for the following reaction is\[1.6\times {{10}^{5}}\]at\[1024\,K.\] \[{{H}_{2}}(g)+B{{r}_{2}}(g)\rightleftharpoons 2HBr(g)\] Find the equilibrium pressure of all gases if 10 bar of\[HBr\]is introduced into a sealed container at 1024 K.

Answer:

The given reaction is: \[{{H}_{2}}(g)+B{{r}_{2}}(g)\rightleftharpoons 2HBr(g);\,\,\,\,\,{{K}_{p}}=1.6\times {{10}^{5}}\] \[\therefore \]\[\underset{{{t}_{\begin{smallmatrix} 0 \\ {{t}_{eq}} \end{smallmatrix}}}}{\mathop{For}}\,\,\,\underset{\begin{smallmatrix} 10 \\ 10-2x \end{smallmatrix}}{\mathop{2HBr(g)}}\,\rightleftharpoons \underset{\begin{smallmatrix} 0 \\ x \end{smallmatrix}}{\mathop{{{H}_{2}}(g)}}\,+\underset{\begin{smallmatrix} 0 \\ x \end{smallmatrix}}{\mathop{B{{r}_{2}}(g)}}\,;\,\,\,{{K}_{p}}=\frac{1}{1.6\times {{10}^{5}}}\] \[{{K}_{p}}=\frac{[{{p}_{{{H}_{2}}}}][{{p}_{B{{r}_{2}}}}]}{{{[{{p}_{HBr}}]}^{2}}}\] \[\frac{1}{1.6\times {{10}^{5}}}=\frac{x\times x}{{{(10-2x)}^{2}}}\] \[\frac{1}{1.6\times {{10}^{4}}}={{\left\{ \frac{x}{10-2x} \right\}}^{2}}\] \[\frac{1}{400}=\frac{x}{10-2x}\] \[10-2x=400x\] \[x=\frac{10}{402}=0.025\] \[\therefore \] \[{{P}_{{{H}_{2}}}}={{p}_{B{{r}_{2}}}}=0.025bar\] \[{{P}_{HBr}}=10-2\times 0.025=9.95bar\]