Which two changes would make this reaction reactant-favored? 2H2 + O2 + 2H20 + energy
A. Increasing the pressure
B. Reducing the temperature
O C. Increasing the temperature
O D. Reducing the pressure

Two changes would make this reaction reactant-favored

C. Increasing the temperature

D. Reducing the pressure

Given

Reaction

2H₂ + O₂ ⇒ 2H₂0 + energy

Required

Two changes would make this reaction reactant-favored

Solution

The formation of H₂O is an exothermic reaction (releases heat)

If the system temperature is raised, then the equilibrium reaction will reduce the temperature by shifting the reaction in the direction that requires heat (endotherms). Conversely, if the temperature is lowered, then the equilibrium shifts to a reaction that releases heat (exothermic)  

While on the change in pressure, then the addition of pressure, the reaction will shift towards a smaller reaction coefficient  

in the above reaction: the number of coefficients on the left is 3 (2 + 1) while the right is 2

As the temperature rises, the equilibrium will shift towards the endothermic reaction, so the reaction shifts to the left towards H₂ + O₂( reactant-favored)

And reducing the pressure, then the reaction shifts to the left H₂ + O₂( reactant-favored)⇒the number of coefficients is greater

i believe the answer is true.