Question

The rate of decomposition of `NH_(3)` on platinum surface is zero order. What are rate of production of `N_(2)` and `H_(2)` if `k=2.5 xx 10^(-4)Ms^(-)`?

Answer 1

`2N_(3)to N_(2)+3H_(2)`
The reaction rate may be given as:
`=-1/2(d[NH_(3)])/(dt)= (d[N_(2)])/(dt) = 1/3 (d[H_(2)])/(dt)`
According to available data,
`(d[NH_(3)])/(dt) = k = 2.5 xx 10^(-4)ms^(-1)`.
`(d[N_(2)])/(dt) = 2.5 xx 10^(-4)ms^(-1)`
`(dH_(2))/(dt) = 3 xx 2.5 xx 10^(-4) ms^(-1)= 7.5 xx 10^(-4)ms^(-1)`