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