Decomposition of `Hl (g)` on Gold surface is zero order reaction. Initially, few moles of `H_(2)` are present in container then which of the following graph is correct ?

Options

A.
B.
C.
D.

Answer

2

Explanation

`2HI(g)rarr H_(2)(g)+I_(2)(g)" "(` zero order `)`
`{:(t=0,a,b,0,bpropP_(H_(2))",initial"),(t=t,a-2x,b+x,,P_(H_(2))prop(b+x)):}`
`rArr " "P_(H_(2))=P_(H_(2)),` initial `+kt(` zero order reaction `)`
So, graph is