Chemistry, 27.08.2019 22:20 chloerodgers56
In a particular experiment at 300°c, [no2] drops from 0.0100 to 0.00650 m in 100 s. the rate of appearance of o2 for this period is m/s.
Answers: 2
![Rate of reaction = -\frac{1}{2} \frac{\Delta [NO_2]}{\Delta t} =\frac{1}{2} \frac{\Delta [NO]}{\Delta t}=\frac{1}{1} \frac{\Delta [O_2]}{\Delta t}](/tpl/images/0203/6913/27878.png)
![\frac{\Delta [NO_2]}{\Delta t} = Rate of disappearance of NO_2](/tpl/images/0203/6913/0404d.png)
![\frac{\Delta [NO]}{\Delta t} = Rate of appearance of NO](/tpl/images/0203/6913/a5133.png)
![\frac{\Delta [O_2]}{\Delta t} = Rate of appearance of O_2](/tpl/images/0203/6913/6255f.png)
![\frac{1}{2} \frac{\Delta [NO_2]}{\Delta t} =\frac{1}{1} \frac{\Delta [O_2]}{\Delta t}](/tpl/images/0203/6913/38a87.png)
![\frac{\Delta [O_2]}{\Delta t} = \frac{1}{2} \frac{0.0035}{100}](/tpl/images/0203/6913/42b7d.png)
![\frac{\Delta [O_2]}{\Delta t} = 0.0000175 or 1.75 \times 10^{-5} M/s](/tpl/images/0203/6913/ac98a.png)
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In a particular experiment at 300°c, [no2] drops from 0.0100 to 0.00650 m in 100 s. the rate of appe...
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