Consider the following reaction at equilibrium: C (s) + H2O (g) ↔ CO (g) + H2 (g) Which of the following conditions will decrease the partial pressure of CO? Consider the following reaction at equilibrium: C (s) + H2O (g) CO (g) + H2 (g) Which of the following conditions will decrease the partial pressure of CO? adding a catalyst to the reaction system increasing the pressure of the reaction vessel decreasing the amount of carbon in the system increasing the amount of H2O (g) decreasing the pressure of the reaction vessel SubmitR

Increasing the pressure of the reaction vessel, will decrease the partial pressure on CO.

Explanation:

Consider the reaction C (s) + H2O (g) ↔ CO (g) + H2 (g)

Although there are the same number of moles of reactants and products, one of the reactants is a solid. Thus, there are fewer moles of gaseous reactants than gaseous products, so decreased volume shifts the system toward reactants.

⇒ adding a catalyst to the reaction system will not change anything to the equilibrium, it will only accelerate the reaction

⇒ increasing the pressure of the reaction vessel

If the pressure of the reaction vessel will be increased, the volume will be decreased, this will make the partial pressure on CO decrease as well.

⇒ decreasing the amount of carbon in the system

If the amount of carbon is decreased, it will not change anything to the number of moles, so the it won't affect the partial pressure.

⇒increasing the amount of H2O (g)

If the amount of H2O is increased, it will not change anything to the number of moles, so the it won't affect the partial pressure.

⇒decreasing the pressure of the reaction vessel

If the pressure of the reaction vessel will be decreased, the volume will be increased this will increase the partial pressure on CO as well.

Decreasing the amount of carbon in the system.

Explanation:

Hello,

In this case, by means of the Le Chatelier principle, the first evident aspect is change in number of gaseous moles, which is zero because the subtraction between water's and carbon monoxide's stoichiometric coefficients, are the gaseous species, is zero, therefore, the variation of neither volume nor pressure will affect the equilibrium. Moreover, the catalysts are used to increase the rate of a chemical reaction favoring the product, so it is also discarded. In such a way, the answer is: decreasing the amount of carbon in the system because the removal of reactants decrease the formation of products, therefore less carbon monoxide will be obtained in terms of of a lower partial pressure.

Best regards.