Finding Equilibrium [C] from initial [C] Question Confusion

[betterfuture]

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Consider the reaction for the decomposition of hydrogen disulfide:

2 H2S(g) ⇌2 H2(g) + S2(g) Kc = 1.67 e-7 at 800°C

The initial concentrations are [H2S] = 0.100M, [H2] = 0.100M and [S2] = 0.00M. Find the equilibrium concentration of [S2].

I don't understand how to solve this completely. Any assistance would help. Thanks.

 

[aldol16]

ICE table.

 

[DrFizition]

Based on Le Châtelier's principle you understand the reaction must shift toward the product side to establish equilibrium. Therefore, in your ICE table, your change would be showing loss of reactant and formation of product:

2 H2S(g) ⇌2 H2(g) + S2(g)
I .1 M -> .1 M 0M
C -x +x +x
E.1-x .1+x x

We have our equilibrium concentration of each expressed in terms of x. Remember, it is the equilibrium concentrations that can be plugged into the K expression. The K expression in this reaction is :

K=([H2]^2[S2]) / (H2S)^2

From here we plug in our values and solve for x. You can use the quadratic equation to do this.

K= ([.1+x][x])/(.1-x)

Once you get x, you can plug that back into the E row of ICE to find the concentration of each gas at equilibrium. Good luck

 

[betterfuture]

ICE table.

Yes, I did do the ICE table. Maybe I should have cleared up what my confusion is coming from. My confusion here is how can the hydrogen gas product form and not the sulfide gas? Isn't sulfide on the side of the products. How does it have a [C] of 0?

Based on Le Châtelier's principle you understand the reaction must shift toward the product side to establish equilibrium. Therefore, in your ICE table, your change would be showing loss of reactant and formation of product:

2 H2S(g) ⇌2 H2(g) + S2(g)
I .1 M -> .1 M 0M
C -x +x +x
E.1-x .1+x x

We have our equilibrium concentration of each expressed in terms of x. Remember, it is the equilibrium concentrations that can be plugged into the K expression. The K expression in this reaction is :

K=([H2]^2[S2]) / (H2S)^2

From here we plug in our values and solve for x. You can use the quadratic equation to do this.

K= ([.1+x][x])/(.1-x)

Once you get x, you can plug that back into the E row of ICE to find the concentration of each gas at equilibrium. Good luck

Does the reaction shift to the right because the [C] of S2(g) is 0? Because the problem has a separate requirement of calculating the reaction quotient Q and some of the other Equilibrium problems did not require me to figure out the reaction quotient.

 

[betterfuture]

Sorry, I meant how can hydrogen gas have an inital [C] and not sulfide gas. Yet they give a [C] of hydrogen gas and states sulfide gas has [C] of 0.

 

[DrFizition]

Let me see if I can answer all of your questions:

My confusion here is how can the hydrogen gas product form and not the sulfide gas? Isn't sulfide on the side of the products. How does it have a [C] of 0?

If I am correct, you are asking "How can we have the initial condition of .1 of H2S, .1 of H2 and 0 S2?" This is based on how the scientist set up this initial system. They simply got a container, inserted .1M H2S, and .1M H2 into it. Then they left it there for equilibrium to be established.

Sorry, I meant how can hydrogen gas have an inital [C] and not sulfide gas. Yet they give a [C] of hydrogen gas and states sulfide gas has [C] of 0.

I believe this is answered by the above statement as well.

Does the reaction shift to the right because the [C] of S2(g) is 0? Because the problem has a separate requirement of calculating the reaction quotient Q and some of the other Equilibrium problems did not require me to figure out the reaction quotient.

Yes, the reaction shifts to the right because S2 is 0. The safest way to know which direction the reaction shifts to establish equilibrium is by finding the reaction quotient. You can do it in this problem too (and from what you said, you had to find Q anyway). But, since we know one of the product's concentrations is 0, it's safe to assume that it will shift right. If you found Q, you will see that Q is < K since Q= 0, meaning the reaction will shift to the right.

 

[betterfuture]

So equilibrium problems do not necessarily have to have initial [C] of just reactants? They can have initial [C] of products as well? Or in this case, only of one of the products and not the other product.

Like you stated, I think they manipulated the system to have the reaction shift to form S2 gas because of Le Chatelier's principle. It's just more common to see initial [C] given of just reactants so maybe that is why I was thrown off by the problem.

 

[betterfuture]

Thanks though for your help and @aldol16 too

 

[DrFizition]

So equilibrium problems do not necessarily have to have initial [C] of just reactants? They can have initial [C] of products as well? Or in this case, only of one of the products and not the other product.

Like you stated, I think they manipulated the system to have the reaction shift to form S2 gas because of Le Chatelier's principle. It's just more common to see initial [C] given of just reactants so maybe that is why I was thrown off by the problem.

Yes, they can set up the initial concentrations to anything. It is up to you to determine which direction the reaction then proceeds.

Don't forget that liquids and solids don't affect equilibrium and don't show up in K or Q expressions.

 

[aldol16]

Sorry, I meant how can hydrogen gas have an inital [C] and not sulfide gas. Yet they give a [C] of hydrogen gas and states sulfide gas has [C] of 0.

You're assuming that the reaction starts with only reactants. The point here is that it does not. For instance, you could stick H2 and O2 in a balloon and react it to form water. Or, you could stick H2, O2, and some water vapor in a balloon and react that to get water too. But in the latter case, you would get less product because you already had some in the system to begin with.

 

[betterfuture]

You're assuming that the reaction starts with only reactants. The point here is that it does not. For instance, you could stick H2 and O2 in a balloon and react it to form water. Or, you could stick H2, O2, and some water vapor in a balloon and react that to get water too. But in the latter case, you would get less product because you already had some in the system to begin with.

Yeah, I guess I didn't think about it that way.