A) 0.331 M
B) 0.033 M
C) 0.133 M
D) 1.33 M
Correct Answer: C
Solution :
[c] \[{{Q}_{C}}=\frac{1\times 1}{1\times 1}=1\] |
\[HN{{O}_{3}}\] \[{{Q}_{C}}>{{K}_{C}}\] so reaction will proceed in backward direction: |
\[{{A}_{2}}(g)+{{B}_{2}}(g)\rightleftharpoons {{C}_{2}}(g)+{{D}_{2}}(g)\] |
Conc. Eqm |
\[\frac{1+x}{10}\] \[\frac{1+x}{10}\] \[\frac{1-x}{10}\] \[\frac{1-x}{10}\] |
\[0.25=\frac{{{\left( \frac{1-x}{10} \right)}^{2}}}{{{\left( \frac{1+x}{10} \right)}^{2}}}\] \[\Rightarrow \] \[x=0.333\] |
\[[A{{ }_{2}}(g)]=\frac{1+x}{10}=\frac{1.333}{10}=0.1333\] |
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