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For the decomposition, `N_(2)O_(5)(g) rarr N_(2)O_(4)(g) + 1//2O_(2)(g)`, the initial pressure of `N_(2)O_(5)` is `114 mm` and after `20 sec`, the pressure of reaction mixture becomes `133 mm`. Calculate the rate of reaction in terms of : (a) change in pressure `sec^(-1)` and (b) change in molarity `sec^(-1)`. Given that reaction is carried out at `127^(@)C`.A. `1.25xx10^(-3) atm sec^(-1)`B. `2.5xx10^(-3) atm sec^(-1)`C. `9.5xx10^(-1) atm sec^(-1)`D. `4.75xx10^(-1) atm sec^(-1)` |
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Answer» Correct Answer - B `{:(N_(2)O_(5),to,N_(2)O_(4),+,(1)/(2)O_(2)),(114.p,,p,,(1)/(2)p):}` `P_(T)=114+(P)/(2)=133rArrp=38` `K=(2.303)/(20)"log"((114)/(76))` `K(Pt)=2.07xx10^(-3)`at m `sec^(-1)` |
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