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Question
For the reaction \[\ce{N2O4 (g) ⇌ 2NO2 (g)}\], the value of K is 50 at 400 K and 1700 at 500 K. Which of the following options is correct?
(i) The reaction is endothermic.
(ii) The reaction is exothermic.
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(iv) The entropy of the system increases.
Solution
(i) The reaction is endothermic.
(iii) If \[\ce{NO2 (g)}\] and \[\ce{N2O4 (g)}\] are mixed at 400 K at partial pressures 20 bar and 2 bar respectively, more \[\ce{N2O4 (g)}\] will be formed.
(iv) The entropy of the system increases.
Explanation:
(i) As the value of K increases with increase of temperature and K = Kf / Kb, this means that kf increases, i.e., forward reaction is favoured. Hence, reaction is endothermic.
(iii) At 400 k, `Q = (P_(N^(2)O_2))/(P_(N_(2)O_4)) = (20)^2/2` = 200. Thus, Q > K. Equilibrium will shift backward to form more \[\ce{N2O4}\].
(iv) As reaction is accompanied by increase in the number of moles, entropy increases.
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