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Equilibrium question help!

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Hello,

This question is for the decomposition of hydrogen iodide:

2HI <------> H2 + I2

I managed to answer the question correctly, however i'm not sure if the way I interpreted the question may have been correct, I would appreciate it if you could confirm whether my approach was correct or not. For experiment one, I assumed that the concentration of HI would decrease by x at equilibrium (0.06 - 2x) and both [H2] and [I2] would increase by x at equilibrium (0.00 + x), as I have been given the concentration at equilibrium for H2, I substitute it in x and find the values for the concentrations at equilibrium of reactants and products to calculate the Kc value. For experiment two, the initial concentration of HI is 0.00 and the initial concentrations of both H2 and I2 are 0.04, this indicates that the reverse reaction is initially faster than the forward reaction and so [HI] at equilibrium will increase by x (0.00 + 2x) and both [H2] and [I2] will decrease by x (0.04 - x), I then equate 0.04 to the concentration of HI at equilibrium (0.00 + 2x = 0.04) and calculate the value of x and find the values for the concentrations at equilibrium of reactants and products to calculate the Kc value. Additionally, how can we determine whether the forward or reverse reaction is faster than the other? It seems quite obvious in this example that the reverse reaction is faster considering the concentration of HI is 0.00 mol dm^-3 for experiment two, but sometimes it may not be so obvious, so how can I know? 

Screen Shot 2016-06-04 at 14.52.16.png

Edited by Scienuk

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Your method is correct.
How to tell if forward reaction is faster or slower than the reverse reaction?

We define the reaction quotient Q, that varies for any time in a reaction (as opposed to the equilibrium constant K, which is constant throughout the reaction):

dLUrOZRMkdx1lIU-f-HEiw_m.jpg

When Q<K, there is less product than at equilibrium position and the forward reaction is faster than reverse reaction;

When Q=K, equilibrium has been reached;

When Q>K, the forward reaction is slower than the reverse reaction

In reaction #1,
Q = 0*0/0.06^2 = 0; K = 0.01*0.01 / 0.04^2 = 0.0625

Because Q < K, the forward reaction proceeds faster than the reverse reaction. 

In reaction #2, 
Q is division by 0 and is undefined. However, we can assign Q = infinite in this case because [HI] cannot be negative, so mathematically, we can say the one-sided limit of Q, approaches infinity as [A] approaches 0. K = 0.02*0.02/0.04^2 = 0.25. Infinity > 0.25 so forward reaction is slower than the reverse reaction (This case you shouldn't worry too much because you said you find the result intuitive).

 

Edited by kw0573

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23 minutes ago, kw0573 said:

Your method is correct.
How to tell if forward reaction is faster or slower than the reverse reaction?

We define the reaction quotient Q, that varies for any time in a reaction (as opposed to the equilibrium constant K, which is constant throughout the reaction):

dLUrOZRMkdx1lIU-f-HEiw_m.jpg

When Q<K, there is less product than at equilibrium position and the forward reaction is faster than reverse reaction;

When Q=K, equilibrium has been reached;

When Q>K, the forward reaction is slower than the reverse reaction

For example for reaction #1,
Q = 0*0/0.06^2 = 0; K = 0.01*0.01 / 0.06^2 = 0.028

Because Q < K, the forward reaction proceeds faster than the reverse reaction. 

What about the second experiment? how am I supposed to know that the reverse reaction is faster than the forward reaction using the reaction quotient Q, can you solve it please and explain, i'm slightly confused. Thank you.

Edited by Scienuk

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