For the reaction: $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$,$\Delta H = -24 \ \text{kcal}$ at $427 \ ^\circ\text{C}$ and $200 \ \text{atm}$. Calculate the magnitude of the internal energy change ($\Delta U$ in $\text{kcal}$),if $168 \ \text{g}$ of $N_2$ gas and $30 \ \text{g}$ of $H_2$ gas are allowed to react completely ($100\%$ reaction yield) to form $NH_3$ gas at $427 \ ^\circ\text{C}$ and $200 \ \text{atm}$.

• A
  $-106$
• B
  $-24$
• C
  $-21.2$
• D
  $-120$