Chemistry, 31.07.2019 18:30, jjjoooorrrrddddaannn

Write the balanced chemical equation for the reverse reaction. include physical states for all species.

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answered: alexabdercmur

The balanced equation for the reverse reaction would be

2SO_3 (g) < ----- 2SO_2 (g) + O_2 (g)

Reversible reactions

They are reactions that proceed in opposing directions - that is, reactants to products and vice versa.

The equilibrium system between sulfur dioxide gas, oxygen gas, and sulfur trioxide gas can be represented as:

2SO_2 (g) + O_2 (g) < --- 2SO_3 (g)

Thus, the reverse reaction would be:

2SO_3 (g) < ----- 2SO_2 (g) + O_2 (g)

More on equilibrium reactions can be found here: link

The equilibrium system between sulfur dioxide gas, oxygen gas, and sulfur trioxide gas is given. Write the balanced chemical equation for the reverse reaction. Include physical states for all species.

answered: speedyblackmann

This is a missing part of your question:
The equilibrium system between sulfur dioxide gas, oxygen gas, and sulfur trioxide gas is given.
So you need the equilibrium balanced equation of SO2, O2, SO3 reaction:
First, we will start with the original equation which is not balanced yet (to understand how we get it):
SO2 + O2 ↔ SO3
Here the number of O atom is not equal at the to sides
So we will start to balance our equation by make the number of O atom equal each other on both sides:
So we will start to put 2SO3 instead of SO3
and put 2SO2 instead of SO2 to balance also the S atom on both sides
So we will get this:
2SO2(g) + O2(g) ↔ 2SO3(g) (This is our equilibrium balanced equation)
know we have a number of O atom equals on each side = 6
and the sulfur equals on each side = 2

answered: Guest

what is the average time it took the car to travel 0.25 meters?

⇒ 2.23 seconds

what is the average time it took the car to travel 0.50 meters?

⇒ 3.13 seconds

answered: Guest

Answerthe half-reaction having the greatest tendency to occur at the anode: i₂(aq) + 2e⁻ ⇄ 2i⁻ ; e = +0.53 v====== $\begin{aligned} \text{i}_2\text{\footnotesize (aq)}+2\,\text{e}^{-} & \rightleftharpoons 2\,\text{i}^{-}\text{\footnotesize (aq)}& \text{e} & = +0.53\text{ v} \\ \text{co}^{3+}\text{\footnotesize (aq)} + \text{e}^{-} & \rightleftharpoons \text{co}^{2+}\text{\footnotesize (aq)} & \text{e} & = +1.80\text{ v} \\ \text{o}_2\text{\footnotesize (g)}+4\,\text{h}^{+}\text{\footnotesize (aq)} + 4\,\text{e}^{-} & \rightleftharpoons 2\,\text{h}_2\text{o}\text{\footnotesize(l)}& \text{e} & = +1.23\text{ v} \\ \end{aligned}$ oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be i₂(aq) + 2e⁻ ⇄ 2i⁻ e = +0.53 vbecause its reduction potential is the lowest out of the three.