Question

In hydrogenation reaction at `25^(@)C`, it is observed that hydrogen gas pressure falls from 2 atm to 1.2 atm isn 50 min. Calculate the rate of reaction in molarity per sec. (R=0.0821 litre -atm `"degree"^(-1) "mol"^(-1))`

Answer 1

Rate `=(dp)/(dt)=(2-1.2)/(50xx60)`
`=2.666xx10^(-4) "atm" s^(-1)`
PV=nRT
`(P)/(sec) =((n)/(V)) =(RT)/(sec)`
`[(n)/(V)](1)/(sec)=[(P)/(sec)](1)/(RT)`
Rate (molarity /sec)=`(2.66xx10^(4))/(0.0821xx298)`
`1.09xx10^(-5) "mol litre"^(-1)s^(-1)`