So the question is: Why is the difference between the heat of reaction at constant pressure and that at constant volume equal to $RT$?
First Part: heat of reaction
From the first law of thermodynamics we know
\begin{equation}
\mathrm{d}U = \delta Q + \delta W = \delta Q - p \mathrm{d} V
\end{equation}
where $U$ is the inner energy, $Q$ is the heat, $W$ is the work, $p$ is the pressure and $V$ is the volume.
So, for a reaction at constant volume you get
\begin{equation}
\mathrm{d}U = \mathrm{d} Q \ .
\end{equation}
Now, for a reaction at constant pressure you can invoke the relationship
\begin{align}
\mathrm{d}H &= \mathrm{d}U + p \mathrm{d}V + V \mathrm{d}p \\
&= \delta Q - p \mathrm{d} V + p \mathrm{d}V + V \mathrm{d}p \\
&= \delta Q + V \mathrm{d}p \\
&\overset{\mathrm{d}p = 0}{=} \mathrm{d} Q
\end{align}
where $H$ is the enthalpy.
This gives you the relationships that describe the change in heats released during reactions at constant volume and constant pressure.
The so called heat of reaction is defined as
\begin{equation}
\Delta Q = \biggl( \frac{\partial Q}{\partial \xi} \biggr)
\end{equation}
where $\xi$ is the extent of reaction.
Second Part: temperature dependence
Now using the equality of mixed partials and the definitions of the heat capacities at constant pressure $C_{p} = \Bigl( \frac{\partial H}{\partial T} \Bigr)_{p}$ and constant volume $C_{V} = \Bigl( \frac{\partial U}{\partial T} \Bigr)_{V}$ you get for constant volume ($\mathrm{d}U = \mathrm{d} Q$):
\begin{align}
\biggl(\frac{\partial \Delta Q}{\partial T} \biggr)_{V} = \Biggl(\frac{\partial \Delta U}{\partial T} \Biggr)_{V} &= \left( \frac{ \Bigl( \frac{\partial U}{\partial \xi} \Bigr)_{T, V}}{\partial T} \right)_{V} = \Biggl( \frac{\partial^{2} U}{\partial \xi \partial T} \Biggr)_{V} \\
&= \Biggl( \frac{\partial^{2} U}{\partial T \partial \xi} \Biggr)_{V} = \left( \frac{ \Bigl( \frac{\partial U}{\partial T} \Bigr)_{V} }{\partial \xi} \right)_{T} = \Biggl(\frac{\partial C_{V}}{\partial \xi} \Biggr)_{T, V} = \Delta C_{V}
\end{align}
and for constant pressure ($\mathrm{d}H = \mathrm{d} Q$):
\begin{align}
\biggl(\frac{\partial \Delta Q}{\partial T} \biggr)_{V} = \Biggl(\frac{\partial \Delta H}{\partial T} \Biggr)_{p} &= \left( \frac{ \Bigl( \frac{\partial H}{\partial \xi} \Bigr)_{T, p}}{\partial T} \right)_{p} = \Biggl( \frac{\partial^{2} H}{\partial \xi \partial T} \Biggr)_{p} \\
&= \Biggl( \frac{\partial^{2} H}{\partial T \partial \xi} \Biggr)_{p} = \left( \frac{ \Bigl( \frac{\partial H}{\partial T} \Bigr)_{p} }{\partial \xi} \right)_{T} = \Biggl(\frac{\partial C_{p}}{\partial \xi} \Biggr)_{T, p} = \Delta C_{p}
\end{align}
where $T$ is the temperature and $\Delta C_{i}$ is change in the molar heat capacities of the reactants and products during the reaction at constant $i$ (where $i$ is either $p$ or $V$):
\begin{align}
\Delta C_{i} = \sum_{k} \nu_{k} c_{i_{k}}
\end{align}
where $\nu_{k}$ and $c_{i_{k}}$ are the stochiometric coefficient and the molar heat capacity at constant $i$ of the $k^{\text{th}}$ component in the reaction, respectively.
When you integrate the equations for the temperature dependences of the reaction heats you get:
\begin{align}
\text{constant volume: } &\int \biggl(\frac{\partial \Delta Q}{\partial T} \biggr)_{V} \, \mathrm{d} T = \left( \Delta Q \right)_{V} = \int \Delta C_{V} \, \mathrm{d} T \\
\text{constant pressure: } &\int \biggl(\frac{\partial \Delta Q}{\partial T} \biggr)_{p} \, \mathrm{d} T = \left( \Delta Q \right)_{p} = \int \Delta C_{p} \, \mathrm{d} T \ .
\end{align}
So, the difference of the reaction heats at constant volume and constant pressure would be:
\begin{align}
\left( \Delta Q \right)_{p} - \left( \Delta Q \right)_{V} = \int \left( \Delta C_{p} - \Delta C_{V} \right) \, \mathrm{d} T \ .
\end{align}
Third Part: ideal gases
Now, why would this boil down to $RT$? This works when you consider reactions that involve only ideal gases. For ideal gases you have (see here):
\begin{align}
c_{p} - c_{V} = R \ .
\end{align}
So, the integral simplifies to
\begin{align}
\int \left( \Delta C_{p} - \Delta C_{V} \right) \, \mathrm{d} T &= \int \sum_{j} \nu_{j} \bigl( \underbrace{ c_{p_{j}} - c_{V_{j}} }_{= \, R} \bigr) \, \mathrm{d} T \\
&= R \int \sum_{j} \nu_{j} \, \mathrm{d} T \\
&= R \sum_{j} \nu_{j} \int \mathrm{d} T \\
&= R T \sum_{j} \nu_{j}
\end{align}
where $\sum_{j} \nu_{j}$ will be some rational number but not necessarily equal to 1 - it depends on the reaction.
This is basically the result you were looking for although the reaction heat difference will only be proportional to $RT$ with the proportionality constant being $\sum_{j} \nu_{j}$:
\begin{align}
\left( \Delta Q \right)_{p} - \left( \Delta Q \right)_{V} = RT \sum_{j} \nu_{j} \ .
\end{align}
