Two changes would make this reaction reactant-favored
C. Increasing the temperature
D. Reducing the pressure
<h3>Further explanation</h3>
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
Answer:
rate= k[A]²[B]²[C]
Explanation:
When concentration of A is increased two times ,keeping other's concentration constant , rate of reaction becomes 4 times .
So rate is proportional to [A]²
When concentration of B is increased two times , keeping other's concentration constant,rate of reaction becomes 4 times.
So rate is proportional to [B]²
When concentration of C is increased two times , keeping other's concentration constant, rate of reaction becomes 2 times.
So rate is proportional to [C]
So rate= k[A]²[B]²[C]
Answer:
As potassium is larger than sodium, potassium's valence electron is at a greater distance from the attractive nucleus and is so removed more easily than sodium's valence electron. As it is removed more easily, it requires less energy, and can be said to be more reactive.
Explanation:
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Electric hidro because he has the highest osmolarity
Do you mean 4+7 if yes it’s 11
