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In the following equilibrium, N_2O_4(g)to2NO_2(g) when 5 moles of each is taken and the temperature is keptat 298K, the total pressure was found to be 20 bar. Given : /_\G_f^0(N_2O_4)=100kJ,/_\G_f^0(NO_2)=50kJ Find /_\G for the reaction at 298K |
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Answer» <P> Solution :For the given reaction, `/_\G^0=2/_\G_f(NO_2)^0-/_\G_(f(N_2O_4)^0``=(2xx50-100)kJ=0` We know, `/_\G^0=-RTinK_p` or `0=-RTInK_p` `:.K_p=1` In the mixture, total NUMBER of MOLES of `N_2O_4` and `NO_2=(5+5)mol=10mol` So, in the mixture, `p_(N_2O_4)=5/(10)xx20"bar"=10"bar"` `p_(NO_2)=5/(10)xx20"bar"=10"bar"` `:.Q_p=(p_(NO_2)^2)/(p_(N_2O_4)^2)=((10^2)/(10))`bar=10bar `Q_pgtK_p`, reaction will occur to a greater extent towards LEFT. |
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