For each of the following reactions, express the given rate of change of concentration of the reactants or Products in that reaction:
a.`H_(2)O_(2) + 2H^(o+) + 3l^(ɵ) rarr I_(3)^(ɵ) + 2H_(2)O`, `(-d[I^(ɵ)])/(d t) = ?`, `(-d[H^(o+)])/(d t) ? `
(b) `16H^(o+) + 2Mn_(4)^(ɵ) + 101^(ɵ) rarr 2Mn^(2+) + 8H_(2)O + 5I_(2)`, `(-d[MNO_(4)^(ɵ)])/(d t) = ?`
( c) `4NH_(3) + 5O_(2) rarr 4NO_(2) + 6H_(2)O` , `(-d[NH_(3)])/(d t) = ?`

Explanation

(a) `H_(2)O_(2) + 2H^(o+) + 3I^(ɵ) rarr I_(3)^(ɵ) + 2H_(2)O`
The equality in this case is
`(-d[H_(2)O_(2)])/(d t) = (1)/(2) (-d[H^(o+)])/(d t) = (1)/(3) (-d[I^(ɵ)])/(d t) = (d[I_(3)^(ɵ)])/(d t)`
`= (1)/(2) (d[H_(2)O])/(d t)`
So `(-d[I^(ɵ)])/(d t) = 3(-d[H_(2)O_(2)])/(d t) = -(3)/(2)(-d[H^(o+)])/(d t)`
`= 3 (d[I_(3)^(ɵ)])/(d t) = (3)/(2) (d[H_(2)O])/(d t)`
(b) `16H^(o+) + 2MnO_(4)^(ɵ) + 101^(ɵ) rarr 2Mb^(2+) + 8H_(2)O + 5I_(2)`
The equality in this case is:
`(1)/(16) (-d[H^(o+)])/(d t) = (1)/(2) (-d[MnO_(4)^(ɵ)])/(d t) = (1)/(10) (-d[I^(ɵ)])/(d t)`
`= (1)/(2) (d[Mn^(2+)])/(d t) = (1)/(8)(d[H_(2)O])/(d t) = (1)/(5) (-d[I_(2)])/(d t)`
So `(-d[MnO_(4)^(ɵ)])/(d t) = (1)/(8) (-d[H^(o+)])/(d t) = (1)/(5) (-d[I^(ɵ)])/(d t)`
`= (d[Mn^(2+)])/(d t) = (1)/(4) (d[H_(2)O])/(d t) = (2)/(5) (d[I_(2)])/(d t)`
( c) `4NH_(3) + 5O_(2) rarr 4NO_(2) + 6H_(2)O`
The equality in this case is:
`(1)/(4) (-d[NH_(3)])/(d t) = (1)/(5) (-d[O_(2)])/(d t) = (1)/(4) (d[NO_(2)])/(d t) = (1)/(6) (d[H_(2)O])/(d t)`
So, `(-d[NH_(3)])/(d t) = (4)/(5) (-d[O_(2)])/(d t) = (d[NO_(2)])/(d t) = (2)/(3) (d[H_(2)O])/(d t)`