# Effect of temperature and Le Chatelier's principle

The question $$\ce{2NO(g) + H_2(g) <=> N_{2}O(g) + H_{2}O} + \pu{351kJ}$$ If the temperature is increased in the reaction above, which of the following occurs?

A. The concentration of $\ce{H_{2}O}$ increases
B. The concentration of $\ce{N_{2}O}$ decreases

My teacher said the answer is B; however, I thought that since the reverse reaction is endothermic, adding more heat would produce more $\ce{2NO}$ and $\ce{H_2}$ which will cause an increase in pressure. But in order to maintain the same pressure, the system needs to decrease the number of moles so it favors the forward reaction because there are 3 moles of the reactants and 2 moles of the products. Did my teacher make a mistake? If he did not can you please explain why he is correct.

Your teacher is right, let's see why. First off, here's our reaction $$\ce{2 NO + H2 -> N2O + H2O}\qquad \Delta H = -351\ \mathrm{kJ/mol}$$ 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 $\mathrm{\frac{[Products]}{[Reactants]}~>1}$. 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.

• So basically since $\ce{N2O (g) + H2O + 351~kilojoules -> 2NO(g) + H2(g)}$ adding more energy will be causing the endothermic reaction to occur more? Jun 3 '14 at 0:31
• Yes, at very low temperatures an equilibrium will favor the thermodynamically most stable side of the equilibrium the most. As you increase the temperature the relative difference in stability between reactants and products becomes less significant and the equilibrium will begin to approach 1
– ron
Jun 3 '14 at 0:51

which will cause an increase in pressure. But in order to maintain the same pressure, the system needs to decrease the number of moles so it favors the forward reaction because there are 3 moles of the reactants and 2 moles of the products.

This is a frequent misconception about Le Chatelier's principle. Pressure only changes the equillbrium if pressure changes in a way that changes concentrations, for example by changing volume or adding more of a reactant or product. If pressure changes in a way that doesn't change concentrations, for example by adding a noble gas to the mixture, or in this case changing temperature, equillibrium is unchanged.

When temperature changes, the equillibrium constant itself changes, according to the Van't Hoff Equation. For exothermic reactions, increasing temperature favors the reverse reaction. For endothermic reactions, increasing temperature favors the forward reaction.

• So the equilibrium constant is governed by partial pressures, not total pressure. Jun 3 '14 at 16:48
• yes, the equillibrium constant is defined as a function of partial pressures or concentrations for idealized situations, fugacity or activity for real situations Jun 3 '14 at 17:00

I'm not sure if you could actually predict what would happen if you raise the temperature of the reaction system. That would change the equilibrium constant K for this equation.

Given that you don't know what the concentration/pressure of each system component is, there is the possibly that K could be adjusted so that it equals Q when you raise the temperature for the "reaction." That would make both answers incorrect since at equilibrium there is no net change in system composition.

Tell your teacher that. He's being ambiguous by not specifying if we're raising the temperature of the reaction system, or whether we're simply heating up the reactants or what.

This example illustrates why your teacher is ambiguous:

$\ce{flour~+~milk~+~eggs ->cake}$

If you heat up the entire reaction system, then you'd have a warm cake, and a warm bag of flour, some warm milk, and some warm eggs. Or maybe you'd have a warm cake. It depends on how much heat you add. Here's one ambiguity.

If you add heat to the reactants only - i.e. you bake the combination of reactants - then you'd get cake.

If you add heat to the product only, then you'd get a warm cake.

Regardless of however crudely your teacher might have phrased the question, you've thought too much into the question. True, adding heat to the product side will encourage the reverse reaction, but you don't know to what extent and it's best not to speculate what might happen also as a result.

• In a system at equilibrium, how can you heat up just the reactants OR just the products? How can you partition where the energy goes into a system? In your example, no matter how much energy you put in or take out of the system, you will never retrieve your cup of milk from your cake.
– long
Oct 25 '14 at 3:48

For an exothermic reaction, $\frac{\Delta G^0(T)}{T}$ increases with increasing T. Since $\ln K=-\frac{\Delta G^0}{RT}$, K decreases with increasing T for an exothermic reaction at all temperatures (not just at the standard temperature of 298 K).