Which gas is released when Potassium reacts with dilute hydrochloric acid?

Answer

D

Explanation

`k_(eq) = (k_(1))/(k_(-1)) ([A^(**)][A])/([A]^(2))` …(i)
`(d[A^(**)])/(dt) = k_(1)[A]^(2)-k_(-1)[A^(**)][A]-k_(2)[A^(**)]` …(ii)
Since `A^(**)` is activated molecules of reactant, applying steady state approximation for `[A^(**)]`,
i.e., `(d[A^(**)])/(dt) = 0` in Eq. (ii), we get
`k_(1)[A]^(-2)-k_(-1)[A^(**)][A]-k_(2)[A^(**)]=0`
`[A^(**)] = (k_(1)[A]^(2))/(k_(-1)[A]+k_(2))`