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2N(2)O(5) rarr 4NO(2) + O(2) If (-d[N(...

`2N_(2)O_(5) rarr 4NO_(2) + O_(2)`
If `(-d[N_(2)O_(5)])/(dt) = k_(1)[N_(2)O_(5)]`
`(d[NO_(2)])/(dt) = k_(2)[N_(2)O_(5)]`
`(d[O_(2)])/(dt) = k_(3)[N_(2)O_(5)]`
What is the relation between `k_(1), k_(2)`, and `k_(3)`?

Text Solution

Verified by Experts

`(-d[N_(2)O_(5)])/(2 dt) = (d[NO_(2)])/(4 dt) = (-d[O_(2)])/(dt)`
`(k_(1)[N_(2)O_(5)])/(2) = (k_(2)[N_(2)O_(5)])/(4) = (k_(3)[N_(2)O_(5)])/(1)`
Multiply by `4`, we get
`2k_(1) = k_(2) = 4k_(3)`
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