InterviewSolution
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InterviewSolution
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For the dissociation reaction `N_(2)OhArr2NO_(2)(g),` the equilibrium constant `K_(P)` is `0.120` atm at `298 K` and total pressure of system is `2` atm. Calculate the degree of dissociation of `N_(2)O_(4)`. |
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Answer» For the reaction `N_(2)O(g)hArr2NO_(2)(g)` `{:("At equilibrium",,1-alpha,,2alpha),("moles",,,,):}` Let `apha` be the degree of dissociation and P is the total pressure then Total number of moles `=2alpha+1-alpha=1+alpha` `:. P_(N_(2)O)= "moles"` of `N_(2)OxxP_(total)` `=((1-alpha)/(1+alpha))P` and `P_(NO_(2))=((2alpha)/(1+apha))P` `:. K_(P)=((P_(NO_(2))))/(P_(N_(2)O))=([((2alpha)/(1+alpha))P]^(2))/(((1-alpha)/(1+alpha))P)` `=(4alpha^(2)P^(2))/(1+alpha)xx(1+alpha)/((1-alpha)P)=(4alpha^(2)P)/(1-alpha^(2)) ...(1)` Given `K_(P)=0.120 "atm", P=2 "atm"` Subtituting all the values in equation (i), we get `0.120=(4alpha^(2)(2))/((1-alpha^(2)))=(8alpha^(2))/((1-alpha^(2)))` `rArr 0.120(1-alpha^(2))=8alpha^(2)` `:.` Degree of dissociation, `alpha=(0.120/8.12)^(1//2)=0.121` |
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