1.

In the reaction 2N_(2)O_(5) to 4 NO_(2) + O_(2) , initial pressure is 500 atm and rate constant K is 3.38 xx 10^(-5) "sec"^(-1) . After 10 minutes the final pressure of N_(2)O_(5) is

Answer»

490 atm
250 atm
480 atm
420 atm

Solution :`p_(0) = 500 ` atm , `K = (2.303)/(t) "log"_(10) (p_(0))/(p_(t))`
`3.38 xx 10^(-5) = (2.303)/(10 xx 60) "log" (500)/(p_(t))`
or `0.00880 `= log `(500)/(p_(t)) implies (500)/(1.02) = 490` atm .


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