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 : The value of equilibrium constant (K) is, 0.004
Explanation :
First we have to calculate the concentration of 
and,
Now we have to calculate the value of equilibrium constant (K).
The given chemical reaction is:
Initial conc. 1.2 0 0
At eqm. (1.2-2x) 2x x
As we are given:
Concentration of 
at equilibrium = x = 0.1 M
The expression for equilibrium constant is:
![K_c=\frac{[SO_2]^2[O_2]}{[SO_3]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BSO_2%5D%5E2%5BO_2%5D%7D%7B%5BSO_3%5D%5E2%7D)
Now put all the given values in this expression, we get:
Thus, the value of equilibrium constant (K) is, 0.004
Hi.
You can calculate mass percent by dividing the mass by the total mass of the compound & multiplying by 100.
28.8 ÷ (171.2 + 28.8) = 0.144
0.144 · 100 = 14.4
I believe the mass percent of the hydrogen in the compound is 14.4%
~
A) environment is most likely the answer. I'm a bit rusty on my biology though. Hope I helped!
