Explanation:
The given reaction is as follows.
Value of equilibrium constant is given as 
= 4.3 \times 10^{6}[/tex].
Concentration of given species is ![[SO_2]](https://tex.z-dn.net/?f=%5BSO_2%5D)
= 0.010 M; ![[SO_3]](https://tex.z-dn.net/?f=%5BSO_3%5D)
= 10.M; ![[O_2]](https://tex.z-dn.net/?f=%5BO_2%5D)
= 0.010 M.
Formula for experimental value of equilibrium constant (Q) is as follows.
Q = ![\frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BSO_%7B3%7D%5D%5E%7B2%7D%7D%7B%5BSO_%7B2%7D%5D%5E%7B2%7D%5BO_%7B2%7D%5D%7D)
Putting the given concentration as follows.
Q =
Q = 
Q = 
It is known that when Q > 
, then reaction moves in the backward direction.
When Q < , then reaction moves in the forward direction.
When Q = , then reaction is at equilibrium.
As, for the given reaction Q > then it means reaction moves in the backward direction.
Thus, we can conclude that the reaction is moving in the backward direction, that is, right to left to reach the equilibrium.
Answer: The correct answer is -297 kJ.
Explanation:
To solve this problem, we want to modify each of the equations given to get the equation at the bottom of the photo. To do this, we realize that we need SO2 on the right side of the equation (as a product). This lets us know that we must reverse the first equation. This gives us:
2SO3 —> O2 + 2SO2 (196 kJ)
Remember that we take the opposite of the enthalpy change (reverse the sign) when we reverse the equation.
Now, both equations have double the coefficients that we would like (for example, there is 2S in the second equation when we need only S). This means we should multiply each equation (and their enthalpy changes) by 1/2. This gives us:
SO3 —>1/2O2 + SO2 (98 kJ)
S + 3/2O2 —> SO3 (-395 kJ)
Now, we add the two equations together. Notice that the SO3 in the reactants in the first equation and the SO3 in the products of the second equation cancel. Also note that O2 is present on both sides of the equation, so we must subtract 3/2 - 1/2, giving us a net 1O2 on the left side of the equation.
S + O2 —> SO2
Now, we must add the enthalpies together to get our final answer.
-395 kJ + 98 kJ = -297 kJ
Hope this helps!
Answer:
I make not know because im the 7th grade but Im give my worthy guess I now this its endothermic
Explanation:
Answer:
The effects of supercritical CO2 (SC-CO2) on the microbiological, sensory (taste, odour, and colour), nutritional (vitamin C content), and physical (cloud, total acidity, pH, and °Brix) qualities of orange juice were studied. The CO2 treatment was performed in a 1 litre capacity double-walled reactor equipped with a magnetic stirring system. Freshly extracted orange juice was treated with supercritical CO2, pasteurised at 90°C, or left untreated. There were no significant differences in the sensory attributes and physical qualities between the CO2 treated juice and freshly extracted juice. The CO2 treated juice retained 88% of its vitamin C, while the pasteurised juice was notably different from the fresh juice and preserved only 57% of its vitamin C content. After 8 weeks of storage at 4°C, there was no microbial growth in the CO2 treated juice.
