Question

For the reaction below, the enthalpy change is ΔH = –566 kJ/mol. Increasing the temperature of the system will cause which of the following to occur? 2CO (g) + O2 (g) ⇄ 2CO2 (g) View Available Hint(s) For the reaction below, the enthalpy change is ΔH = –566 kJ/mol. Increasing the temperature of the system will cause which of the following to occur? 2CO (g) + O2 (g) ⇄ 2CO2 (g) The partial pressure of CO2 will decrease. The partial pressure of CO2 will increase. The partial pressure of CO will decrease. The partial pressure of O2 will decrease.

Answer 1

The partial pressure of CO₂ will increase

Step-by-step explanation:

• The reversible reaction 2CO (g) + O₂ (g) ⇄ 2CO₂ (g) is exothermic since its enthalpy change is a negative value, ΔH = -566 kJ/mol.
• Exothermic reactions are reactions that release heat energy to the surroundings and thus the energy of products is less than the energy of the reactants.
• According to Le Chatelier's principle, an increase in temperature in the above reaction will favor the reverse reaction. Therefore, the concentration of CO₂ would decrease, while that of CO and O₂ will increase.
• Consequently, the partial pressures of CO and O₂ will increase while that of CO₂ will decrease.

Answer 2

The partial pressure of CO₂ will decrease.

Step-by-step explanation:

The reaction:

2CO (g) + O₂ (g) ⇄ 2CO₂ (g) has a ΔH = –566 kJ/mol. As ΔH<0, the reaction is exothermic.

Le Chatelier's principle says that if a system in chemical equilibrium is subjected to a disturbance it tends to change in a way that opposes this disturbance.

In this case, with increasing of the temperature, the system will produce less heat, doing the equilibrium shifts to the left.

Thus, the partial pressure of both CO and O₂ will increase. And partial pressure of CO₂ will decrease.

I hope it helps!