A) 65%
B) 61.3%
C) 62.5%
D) 64%
Correct Answer: B
Solution :
\[2N{{H}_{3}}\] ⇌ \[{{N}_{2}}+\] \[3{{H}_{2}}\] Initial mole \[a\] 0 0 Mole at equilibrium \[(a-2x)\] \[x\] \[3x\] Initial pressure of \[N{{H}_{3}}\] of a mole = 15 atm at \[{{27}^{o}}C\] The pressure of 'a' mole of \[N{{H}_{3}}=p\] atm at \[{{347}^{o}}C\] \[\therefore \] \[\frac{15}{300}=\frac{p}{620}\] \[\therefore \] \[p=31\]atm At constant volume and at \[{{347}^{o}}C\], mole \[\propto \] pressure \[a\propto 31\] (before equilibrium) \[\therefore \] \[a+2x\,\propto 50\] (after equilibrium) \[\therefore \] \[\frac{a+2x}{a}=\frac{50}{31}\] \[\therefore \] \[x=\frac{19}{62}a\] \[\therefore \] % of \[N{{H}_{3}}\] decomposed \[=\frac{2x}{a}\times 100\] \[=\frac{2\times 19a}{62\times a}\times 100=61.33%\]You need to login to perform this action.
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