Answer:
(a, c, d)
·
Equilibrium constant is increasing with temperature hence, the reaction
will be endothermic.
\[\log \left( \frac{{{K}_{2}}}{{{K}_{1}}}
\right)=\frac{\Delta H}{2.303R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}}
\right]\]
·
\[{{Q}_{p}}=\frac{p_{N{{O}_{2}}}^{2}}{{{p}_{{{N}_{2}}{{O}_{4}}}}}=\frac{{{(20)}^{2}}}{2}=200\]
\[\underset{(200)}{\mathop{{{Q}_{p}}}}\,=\underset{50}{\mathop{{{K}_{p}}}}\,,\]
hence reaction will be fast in backward
direction to form more
\[{{N}_{2}}{{O}_{4}}.\]
·
\[\Delta {{n}_{g}}=2-1\], i.e., \[\Delta {{n}_{g}}>0.\], hence
entropy will increase in the reaction.
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