Question

question 5 (1 point)
give ways that could be used to increase production of $\ce{p2o5}$ in the reaction if all are gases.
$\ce{4 ph3 + 2 h2 + 9 o2 <=> 2 p2o5 + 8 h2o + 25 kj}$
we want the overall reaction to go 1
2 $\ce{ph3}$
3 $\ce{h2}$
4 $\ce{o2}$
5 $\ce{h2o}$
6 the temperature
7 the pressure
add a catalyst since the reaction is no longer at equilibrium
a. increase b. decrease c. no change d. forward → e. reverse ←

Explanation:

To increase \( P_2O_5 \) production, we use Le Chatelier’s principle:

1. Reaction Direction: Need forward (\( D \)) to make more \( P_2O_5 \).
2. Reactants (\( PH_3, H_2, O_2 \)): Increasing (\( A \)) reactants shifts equilibrium forward.
3. Product (\( H_2O \)): Decreasing (\( B \)) \( H_2O \) (a product) shifts equilibrium forward.
4. Temperature: The reaction releases heat (exothermic, \( +25 \, \text{kJ} \)). Decreasing (\( B \)) temperature favors the exothermic (forward) direction.
5. Pressure: Count gas moles: Reactants: \( 4 + 2 + 9 = 15 \); Products: \( 2 + 8 = 10 \). Increasing (\( A \)) pressure shifts to the side with fewer moles (products), favoring forward.

Answer:

1. D. forward →
2. A. increase
3. A. increase
4. A. increase
5. B. decrease
6. B. decrease
7. A. increase