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`CH_(4) (g)` and `O_(2) (g)` react according to the given equation: `CH_(4) (g) + 2O_(2) (g) rarr CO_(2) (g) + 2H_(2) O (l)` Assuming that reaction take placed and goes to completion.If valve is opened then [Assume temperature remains constant `300 K` throughout ] A. `CH_(4)` gas is left after the reactionB. `O_(2)` gas is left the reaction.C. Total pressure is `(1)/(3)` atm after the reaction .D. After the reaction pressure in both the vessels will remain same |
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Answer» Correct Answer - A::C::D `n_(CH_(4)) = (0.5 xx 2)/(R xx 300) = (1)/(300 R)` `n_(O_(2)) = (1 xx 1)/(R xx 300) = (1)/(300 R)` `{:(,CH_(4) (g),+,2O_(2) (g),rarr,CO_(2) (g),+,2H_(2) o (l),),("Mole initially",(1)/(300R),,(1)/(300 R),,-,,-,):}` `O_(2)` is LR `CH_(2)` will be left after the reaction `{:("Mole at",(1)/(600 R),0,(1)/(600 R),-,),("Completion",,,,,):}` Total moles after completion `= n_(CH_(4) + n_(CO_(2))` `= (1)/(600 R) + (1)/(600 R)` `(1)/(300 R)` `P = (nRT)/(V) = (1)/(300 R) xx (R xx 300)/(3) = (1)/(3)` atm |
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