Effect of temperature and Le Chatelier's principle
Your teacher is right, let's see why. First off, here's our reaction $$\ce{2 NO + H2 -&gt; N2O + H2O}\qquad \Delta H = -351\ \mathrm{kJ/mol}$$
So the reaction to the right is exothermic, it gives off heat.  That tells us the products are more stable than the reactants and at equilibrium $\mathrm{\frac{[Products]}{[Reactants]}~&gt;1}$.  Here we could guess that if we heat the reaction (put heat back into the reaction), we will tend to level out the equilibrium thereby pushing the reaction back to the left.

To examine this guess let's work an example.  You were given an enthalpy of reaction, but for a "back of the envelope" calculation, let's run a comparison and assume that the change in entropy will be roughly the same for the two conditions (temperatures) we're going to compare.  With this assumption, we can let $$\Delta G= -RT\cdot\ln K_\text{eq}= -351\ \mathrm{kJ/mol}$$  We know $\Delta G$, $R=8.31\times10^{-3}$ and let's say the data is given for room temperature, so $T=298\ \mathrm K$.  Solving for $K_\text{eq}$ we find $K_\text{eq}=3.6\times10^{61}$.  Now let's let our reaction equilibrate $100^\circ$ above room temperature (now $T=398\ \mathrm K$).  $\Delta G$ and $R$ are fixed, so lets resolve the equation and we now find $K_\text{eq}=1.2\times10^{46}$, indeed heat has pushed our reaction to the left.  The teacher was right.