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The decompoistion of `Cl_(2)O_(7)` at `400 K` in gas phase to `Cl_(2)` and `O_(2)` is a first order reaction. a. After `55 s` at `400 K`, the pressure of `Cl_(2)O_(7)` falls form `0.062` to `0.044 atm`. Calculate `k`. b. Calculate the pressure of `Cl_(2)O_(7)` after `100 s` of decompoistion. |
Answer» As pressuer `prop` concentration, `k=(2.303)/(t)"log"_(10)(P_(i)("initial pressure"))/(P_(t)("pressure after time t"))` `=(2.303)/(55)"log"_(10)(0.062)/(0.044)=6.2xx10^(-3)s^(-1)` (ii) Again applying the first order kinetic eqution, `k=(2.303)/(t)"log"_(10)(P_(i)("intial pressure"))/(P_(t)("pressure after time t"))` `6.2xx10^(-3)=(2.303)/(100)"log"_(10)(0.062)/(P_(t)` `or(6.2xx10^(-3)xx100)/(2.303)="log"0.062-log_(10)(P_(t))` `or 0.2692=log_(10)0.062-0.2692` `or log_(10)(P_(t))=log_(10)0.062-0.2692` `=(bar(2).7924-0.2692)` `P_(t) =0.033` atmoshpere Pressure after 100 sec =0.033 atm |
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