Q10 (5pts)_[EXPLAIN your ANSWER FOR CASE A) B) C) D) E) (NO EXPLANATION OR WRONG

Question

Q10 (5pts)_[EXPLAIN your ANSWER FOR CASE A) B) C) D) E) (NO EXPLANATION OR WRONG EXPLANATION OR PARTIAL EXPLANATION 0 pts). Consider the following reaction at equilibrium: C (s) + H20CO (g) + H2 (g.) Which of the following conditions will decrease the partial pressure of Co? A) decreasing the volume of the reaction vessel B) increasing the volume of the reaction vessel C) decreasing the amount of carbon in the system D) decreasing the pressure of the reaction vessel E) adding a catalyst to the reaction system

Explanation / Answer

For the given reaction :

C (s) + H2O (g) <-----------------> CO (g) + H2 (g)

A) Yes , Decreasing the volume of the reaction vessel will decrease the partial pressure of CO because decreasing volume will lead to decrease in number of moles and the moles are less towards left side and therefore reaction will go to left side and CO will be consumed.

B)  No , Increasing the volume of the reaction vessel will not decrease the partial pressure of CO because increasing volume will lead to increase in number of moles and the moles are more towards left side and therefore reaction will go to right side and CO will be formed.

C) No effect as Carbon is in solid form and it remains almost constant.

D)  No , decreaing the pressure will lead to Increase the volume of the reaction vessel and it will not decrease the partial pressure of CO because increasing volume will lead to increase in number of moles and the moles are more towards left side and therefore reaction will go to right side and CO will be formed.

E) No effect as catalyst will increase the rate of reaction at the same rate on the both side.

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