Answer
$\text{H}_2$ dissociates more than $\text{N}_2$. Value of equilibrium constant for $\text{H}_2$ dissociation is more than that for $\text{N}_2$.
Work Step by Step
From the Tables "Natural logarithmic values of equilibrium constant", the logarithmic values of dissociation for $\text{H}_2$ and $\text{N}_2$ are obtained as,
$\begin{array}{l}\text{For N}_2: \mathrm{N}_{2} \Leftrightarrow 2 \mathrm{N} \text{ We have } \ln K_{p}=-22.359 \\\text{For H}_2: \mathrm{H}_{2} \Leftrightarrow 2 \mathrm{H} \text{ We have } \ln K_{p}=-3.685\end{array}$
Since equilibrium constant is greater for $\text{H}_2$, it dissociates more.
