Is Delta H = Delta U for bomb calorimeter or Delta H = Delta U + Delta n(g)RT?
First and foremost your $-0.5$ should have units of moles (i.e. should be $-0.5~\mathrm{mol}$). Don't leave out units. They're important.


  $$\Delta H = \Delta U + p\Delta V$$


If you were not taught what conditions this equation holds true, shame on your teacher. If you were taught it, you should probably try to remember that this holds for constant pressure.

In general

$$\begin{align}
H &amp;= U + pV \\
H_2 - H_1 &amp;= (U_2 + p_2V_2) - (U_1 + p_1V_1) \\
&amp;= (U_2 - U_1) + (p_2V_2 - p_1V_1) \\
\Delta H &amp;= \Delta U + \Delta (pV)
\end{align}$$

Only under constant pressure $p_2 = p_1 = p$ does this simplify to

$$\begin{align}
\Delta H &amp;= (U_2 - U_1) + (pV_2 - pV_1) \\
&amp;= \Delta U + p(V_2 - V_1) \\
&amp;= \Delta U + p\Delta V \\
\end{align}$$

If this is not true (and think about it, if your volume is constant, then what do you think is happening to the pressure inside the calorimeter?), then you may use the ideal gas approximation

$$p_2V_2 = n_2RT_2;\quad  p_1V_1 = n_1RT_1$$

and if you further assume constant temperature $T_2 = T_1 = T$ (which is probably not true, but it is close enough to the truth since the heat capacity of a calorimeter is usually large leading to a very small difference between $T_2$ and $T_1$), then

$$\begin{align}
\Delta H &amp;= (U_2 - U_1) + (n_2RT - n_1RT) \\
&amp;= \Delta U + (n_2 - n_1)RT \\
&amp;= \Delta U + RT\Delta n
\end{align}$$

The further approximation is that the non-gaseous species do not contribute significantly to the $\Delta(pV)$ term, so $\Delta n$ here can be identified with $\Delta n_g$, the difference in the number of moles of gas.

TL;DR

$$\color{red}{\textbf{Learn when the equations you are using are appropriate to use.}} \\
\large\color{red}{\textbf{Don't just blindly use them!}}$$