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A sample of air consisting mainly of N_2 and O_2 is heated at 2500K till the given equilibrium attains- N_2(g) + O_2 (g) hArr 2NO(g). If the value of equilibrium constant (K_c) at this temperature is 2.1xx10^(-3) and the percentage of number of moles of NO(g) in the above equilibrium mixture is 1.8 %, then what will be the mole fractions of N_2 and O_2 in the sample of air used'? |
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Answer» Solution :SUPPOSE, initial total NUMBER of moles of `N_2` and `O_2` in the sample of air = 1 mol. In the sample, if the initial number of moles of `N_2` is n, then the same for `O_2` is (1-n) mol. Let, the number of moles of `N_2` REACTS at equilibrium=`prop` Initial number of moles : `N_2+O_2 Number of moles at equilibrium: `(n-prop)(1-n-prop)2prop` Total number of moles at equilibrium `=n-prop+1-n-prop+2prop=1` so, mole percentage of NO`=(2prop)/(1_xx100=1.8` `:.prop=9xx10^(-3)` `K_c=([NO]^2)/([N_2][O_2])=(n_(NO)^2)/(n_(N_2)xxn_(O_2))` `=((2xx9xx10^(-3))^2)/((n-9xx10^(-3)(n-9xx10^(-3))=0.154` `:.` n=0.21 so, initial number of moles of `O_2`=0.21 and that of `N_2=1-0.21=0.79`. Hence, in the sample of air, mole FRACTION of `O_2`=0.21 and that of `N_2`=0.79 |
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