The decompoistion of ammonia on platinum surface follows the change
`2NH_(3) rarr N_(2) + 3H_(2)`
(a) What does `(-d[NH_(3)])/(dt)` denote?
(b) What does `(d[N_(2)])/(dt)` and `(d[H_(2)])/(dt)` denote?
( c) If the decompoistion is zero order then what are the rates of Production of `N_(2)` and `H_(2)` if `k = 2.5 xx 10^(-4) M s^(-1)` ?
If the rates obeys `- (d[NH_(3)])/(d t) = (k_(1)[NH_(3)])/(1+k_(2)[NH_(3)])`, what will be the order for decompoistion of `NH_(3)`, if (i) `[NH_(3)]` is every less and (ii) `[NH_(3)]` is very high? (`k_(1)` and `k_(2)` are constants)

Explanation

(a) The rate of decompoistion of `NH_(3)`.
(b) The rate of formation of `N_(2)` and rate of formation of `H_(2)`.
( c) `-(1)/(2) (d[NH_(3)])/(dt) = (d[N_(2)])/(dt) = (1)/(3)(d[H_(2)])/(d t) = k[NH_(3)]^(0)`
`:. (d[N_(2)])/(dt) = k = 2.5 xx 10^(-4) M s^(-1)`
`(d[H_(2)])/(dt) = 3k = 3 xx 2.5 xx 10^(-4) = 7.5 xx 10^(-4) M s^(-1)`
(d) `-(d[NH_(3)])/(dt) = (k_(1)[NH_(3)])/(1+k_(2)[NH_(3)])`
If `[NH_(3)]` is very law, then `k_(2)[NH_(3)]` can be neglected compared to unity and we have,
`(-d[NH_(3)])/(dt) = k_(1)[NH_(3)]`, i.e., first order reaction w.r.t. `NH_(3)`.
If `[NH_(3)]` is very high, we may neglect unity compared to `k_(2)[NH_(3)]` and we have.
`(-d[NH_(3)])/(dt) = (k_(1)[NH_(3)])/(k_(2)[NH_(3)]) = (k_(1))/(k_(2)) = k`
It can be seen that the rate is constant, a characteristic of zero order reaction.