Answer:
\[\underset{{{t}_{\begin{smallmatrix}
0 \\
{{t}_{eq}}
\end{smallmatrix}}}}{\overset{{}}{\mathop{{}}}}\,\,\,\,\,\underset{\begin{smallmatrix}
0.087 \\
(0.087-2x)
\end{smallmatrix}}{\mathop{2NO(g)}}\,\,+\underset{\begin{smallmatrix}
0.0437 \\
(0.0437-x)
\end{smallmatrix}}{\mathop{B{{r}_{2}}(g)}}\,\rightleftharpoons
\underset{\begin{smallmatrix}
0 \\
2x
\end{smallmatrix}}{\mathop{2NOBr(g)}}\,\]
Given,
\[2x=0.0518\]
x = 0.0259 mol
Equilibrium
amount of
\[B{{r}_{2}}\,=(0.0437-0.0259)mol\]
= 0.0178 mol
Equilibrium amount
of \[NO\text{ }=\text{ }0.087\text{ }-2x\]
=
0.087 - 0.0518 = 0.0352 mol
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