inorganic chemistry - How does H2SO4 dissociate?

Apologies for this extremely basic question, I'm just beginning with Chemistry so please don't be too harsh on me.

My book says that sulfuric acid, $\ce{H2SO4}$, dissociates in its ions following this reaction: $$\ce{H2SO4 -> H2^+ + SO4^{2-}}$$

My question is, why can't the dissociation reaction happen like this: $$\ce{H2SO4 -> 2H^+ +SO4^{2-}}$$

I know hydrogen is a diatomic gas, but here I don't know if H will dissociate as a gas or as a liquid (since $\ce{H2SO4}$ is a liquid, not a gas).

I'm trying to learn, thank you for your understanding and your time.

Answer

$\ce{H2SO4}$ is one of common strong acids, meaning that $\ce{K_{a(1)}}$ is large and that its dissociation even in moderately concentrated aqueous solutions is almost complete.

Arrhenius dissociation:

$$\ce{H2SO4 <=> H+ + HSO4-}~~~~~~~~~~\ce{K_{a(1)}}=\ce{large}$$

Brønsted-Lowry Dissociation:

$$\ce{H2SO4 + H2O <=> H3O+ + HSO4-}~~~~~~~~~~\ce{K_{a(1)}}=\ce{large}$$

This accounts for the vast majority of protons donated by the acid. However, since it is diprotic, you may want to take into account the second dissociation, which is technically weak but has a larger $\ce{K_a}$ than many weak acids.

Arrhenius 2nd Dissociation:

$$\ce{HSO4- <=> H+ + {SO_4}^2-}~~~~~~~~~~\ce{K_{a(2)}}=1.2\times10^{-2}$$

Brønsted-Lowry 2nd Dissociation:

$$\ce{HSO4- + H2O <=> H3O+ +{SO_4}^2-}~~~~~~~~~~\ce{K_{a(2)}}= 1.2\times10^{-2}$$

This second dissociation may need to be taken into account for some calculations, but it is negligible in concentrated solutions.