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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)` isA. 490 atmB. 250 atmC. 480 atmD. 420 atm |
Answer» Correct Answer - a `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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