
Anonymous A. answered 03/20/24
Synthetic Organic Chemist with Experience Tutoring Chemistry
If the combination of sulfur dioxide and oxygen has a negative delta H, then it is an exothermic reaction and heat can be thought of as a product much like SO3. Since this is an equilibrium reaction, LeChatelier's principle will apply. If you lower the temperature of the reaction, you are effectively decreasing the amount of heat present and, so decreasing the amount of one of the products. To re-establish equilibrium, the reaction will push itself forward to form even more SO3, so answer A is incorrect.
If nitrogen oxides were needed as additional catalysts and constantly formed as choice C suggests, then there would be less oxygen available for the main reaction. Since it's an equilibrium reaction, a decrease in one of the reactants' concentrations would lead to a shift of the reaction backwards to starting material, which is not a desired outcome.
Choice D is nonsensical because it's saying once the reaction starts and releases heat at 450 C, the temperature of the catalyst bed rises to 600 C at which the point the reaction begins. But the reaction already began. It doesn't need to be at 600 C if it is already releasing heat at 450 C.
Choice B is the only correct choice. Metal catalysts are very sensitive to changes in temperature and oftentimes gaseous equilibrium reactions like this one and also the one in the Haber-Bosch process require temperatures in excess of 400 C for the metallic catalyst to work. From a thermodynamic perspective, decreasing the temperature WOULD create more product according to LeChatelier's principle but if there is not enough energy to overcome a reaction's intrinsic kinetic barrier, nothing will happen.