The reaction 2SO + O2 �¨ 2SO3 is exothermic. (a) How will a rise in temperature affect the yield of SO3 in an equilibrium mixture of the three gases? (b) Will an increase in pressure raise or lower the yield? (c) In what possible way can the speed of the reaction be increased at moderate temperatures?
2SO + O2 --> 2SO3 + heat
Le Chatelier's Principle tells us, in simple terms, that a system in equilibrium will shift so as to undo what we've done. So if the heat is evolved and we increase the temperature the reaction will shift so as to use up the heat we've added. Therefore, the reaction will shift to the ...... (SO3) will ...... while SO and O2 will .....
b) An increase in pressure will make it shift to the side with fewer moles. There are three moles on the left and 2 on the right. So it will shift to the ..... produce (more/less) SO3 which will (increase/decrease) the yield.
To answer these questions, we need to understand the principles of Le Chatelier's principle and the factors that affect reaction rates.
(a) How will a rise in temperature affect the yield of SO3 in an equilibrium mixture of the three gases?
According to Le Chatelier's principle, when the temperature of a system at equilibrium is increased, the equilibrium will shift in the endothermic direction to counteract the rise in temperature. In this case, the reaction is exothermic, which means it releases heat. Thus, an increase in temperature will shift the equilibrium in the reverse direction (towards the reactants) to consume and offset the excess heat. As a result, the yield of SO3 will decrease.
(b) Will an increase in pressure raise or lower the yield?
In this reaction, pressure does not affect the yield of SO3 because there is no change in the number of moles of gas. The balanced equation shows that two moles of SO2 and one mole of O2 react to form two moles of SO3. The total number of moles before and after the reaction remains the same, so changes in pressure will not alter the equilibrium position or the yield.
(c) In what possible way can the speed of the reaction be increased at moderate temperatures?
To increase the speed of the reaction at moderate temperatures, we can utilize the following methods:
1. Increase the concentration of reactants: By adding more SO2 and O2, the chances of successful collisions between the reactant molecules increase, leading to a higher reaction rate. This can be accomplished by adding an excess of either or both reactants.
2. Increase surface area or catalysts: Breaking down the reactants into smaller particles increases the surface area available for collisions, thereby enhancing the reaction rate. Additionally, adding a catalyst can provide an alternate reaction pathway with lower energy barriers, speeding up the reaction.
3. Increase temperature: Increasing the temperature not only affects the equilibrium (as mentioned in part (a)), but it also boosts the kinetic energy of the particles, causing them to move faster and collide more frequently. This leads to an increased reaction rate.
Note: While an increase in temperature can increase the reaction rate, it may also favor the reverse reaction at equilibrium, reducing the yield of the desired product (as discussed in part (a)).