question_answer 121)

    In the following questions two or more options may be correct. For the reaction, \[{{N}_{2}}{{O}_{4}}(g)\underset{{}}{\leftrightarrows}2N{{O}_{2}}(g),\] the value of K is 50 at 400 K and 1700 at 500 K. Which of the following options is correct? (a) The reaction is endothermic (b) The reaction is exothermic (c) If \[N{{O}_{2}}(g)\] and \[{{N}_{2}}{{O}_{4}}(g)\] are mixed at 400 K at partial pressures 20 bar and 2 bar respectively, more \[{{N}_{2}}{{O}_{4}}(g)\]will be formed (d) The entropy of the system increases

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.