Question

An equimolar mixture of \(\mathrm{CO}\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) reacts to form an equilibrium mixture of \(\mathrm{CO}_{2}, \mathrm{CO}, \mathrm{H}_{2} \mathrm{O}\), and \(\mathrm{H}_{2}\) at \(1727^{\circ} \mathrm{C}\), 1 atm. (a) Will lowering the temperature increase or decrease the amount of \(\mathrm{H}_{2}\) present? Explain. (b) Will decreasing the pressure while keeping the temperature constant increase or decrease the amount of \(\mathrm{H}_{2}\) present? Explain.

Short answer

(a) Lowering the temperature decreases the amount of \(\mathrm{H}_{2}\). (b) Decreasing the pressure has no effect.

Step-by-step solution

Write the Balanced Chemical Equation

The reaction can be represented as: \[ \mathrm{CO} + \mathrm{H}_{2}\mathrm{O}(\mathrm{g}) \leftrightarrow \mathrm{CO}_{2} + \mathrm{H}_{2} \] Identify the reactants and products.

Apply Le Chatelier's Principle for Temperature

Le Chatelier's Principle states that if a system at equilibrium is disturbed, the system will adjust to counteract the disturbance and restore equilibrium. Since the reaction is endothermic, lowering the temperature will shift the equilibrium towards the exothermic direction (reactants) to produce heat. Therefore, the amount of \(\mathrm{H}_{2}\) will decrease.

Apply Le Chatelier's Principle for Pressure

Le Chatelier's Principle also applies to changes in pressure. For the given reaction, the number of moles of gas does not change (1 mole of reactants produces 1 mole of products). Therefore, decreasing the pressure will have no effect on the amount of \(\mathrm{H}_{2}\) present.

Key concepts

chemical equilibrium

Chemical equilibrium occurs in a reversible reaction when the rate of the forward reaction equals the rate of the backward reaction. At this point, the concentrations of the reactants and products remain constant over time. In our reaction The backward reaction produces For the mixture to be at equilibrium, the rates of these two processes become equal, leading to constant concentrations of all substances involved. It is important to recognize that equilibrium does not mean the reactants and products are equal in concentration, but rather that their concentrations do not change as long as the conditions remain constant. Le Chatelier's Principle gives us a way to predict how the equilibrium will shift if we disturb the system.

endothermic reaction

The reaction between CO and H2O to form CO2 and H2 is an example of an endothermic reaction. Remember that an endothermic reaction absorbs heat from its surroundings. This means that heat acts like a reactant in this type of reaction. When the temperature of an endothermic reaction like this is increased, more heat is added to the system. To counteract this and restore equilibrium, the system shifts towards the products, producing more CO2 and H2. Conversely, lowering the temperature of an endothermic reaction causes the equilibrium to shift towards the reactants. By reducing heat, the system compensates by making more In our given reaction, lowering the temperature will shift the equilibrium to the left side, decreasing the amount of H2 present.

pressure and temperature effects

Le Chatelier's Principle not only applies to changes in temperature but also to changes in pressure. When dealing with gaseous reactions, changing the pressure can have significant effects on the equilibrium.For reactions where the number of gas moles differs between reactants and products, altering the pressure will shift the equilibrium towards the side with fewer moles of gas when pressure is increased, and towards the side with more moles when pressure is decreased.In the reaction Because the number of gas moles on both sides is equal, a change in pressure will not affect the equilibrium position. Hence, decreasing the pressure in this scenario will not change the amount of H2 present in the equilibrium mixture. To summarize the effects:

• Lowering temperature: shifts equilibrium towards reactants in endothermic reactions.
• Decreasing pressure: has no effect if the number of gas molecules remains the same.