In which of the following gas-phase reactions would the equilibrium yield increase with increasing pressure?
The question tells us that all the reactants and all the products are gases. The ideal gas law relates the pressure and volume of an ideal gas with the amount of substance in moles, the ideal gas constant, and the temperature. If volume and temperature are constant, then the pressure in ideal gas is directly proportional to the amount of substance. This property will come in handy a little later on. The equilibrium yield is the percentage yield at equilibrium. So it’s the amount of product we achieve at equilibrium relative to what we would achieve if we had 100 percent yield.
The question is why would increasing the pressure affect the equilibrium yield? Le Chatelier’s principle tells us that a position of an equilibrium will shift to counteract a change. Here, our change isn’t increase in the pressure. So in this case, Le Chatelier’s principle tells us that a position of equilibrium will shift in favour of the reaction that reduces the amount of substance, the number of moles. In our first reaction, we have three moles of reactant molecules producing two moles of product molecules. So the forward reaction in this equilibrium decreases the amount of substance, while the reverse reaction increases it.
An increase in pressure for this reaction will favour whichever reaction reduces 𝑛. That’s the forward reaction. That’s because the forward reaction will lead to a reduction in pressure counteracting the increasing pressure exerted on the system. So the equilibrium position shifts to the right as the pressure increases. Since shifting to the right means that we get more product, our equilibrium yield increases. Therefore, this is our correct answer. We’ve demonstrated that the increase in pressure shifts the equilibrium to the right because the number of particles in the products is less than that for the reactants.
Just to be safe, let’s have a look at the other options. For the second equilibrium, two particles turn into four. The amount of substance increases due to the forward reaction. Therefore, an increase in pressure will lower the equilibrium yield. The same is true for the third equilibrium, while for the fourth there is no change in the number of particles on both sides of the equation.
So, therefore, the change in pressure will not affect the equilibrium yield. And the same is true for the last equilibrium. Therefore, for all these four options, since increasing the pressure will either lower the equilibrium yield or keep it the same, they are incorrect answers. Therefore, of the five options given, the only gas-phase reaction where the equilibrium yield will increase with increasing pressure is two 2NO₂ plus CH₃OH in equilibrium with CH₃ONO plus HNO₃.