Equilibrium vs. Non-Equilibrium

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Delaney Smith 1C
Posts: 52
Joined: Thu Feb 28, 2019 12:15 am

Equilibrium vs. Non-Equilibrium

Postby Delaney Smith 1C » Sun Jan 12, 2020 4:19 pm

How do you know if a reaction is at equilibrium or not at equilibrium? Will the question state specifically to solve for K or Q or do you have to know just by looking at the reaction whether you're solving a reaction at equilibrium or not at equilibrium?

Radha Patel 4I
Posts: 97
Joined: Tue Feb 12, 2019 12:15 am

Re: Equilibrium vs. Non-Equilibrium

Postby Radha Patel 4I » Sun Jan 12, 2020 4:27 pm

I think the question will provide a value for K then you would have to find the value of Q or it will provide Q and you would have to find K. And then you would use the values of K and Q to determine if they are at equilibrium if both K and Q are the same.

Alice Chang 2H
Posts: 101
Joined: Fri Aug 30, 2019 12:18 am

Re: Equilibrium vs. Non-Equilibrium

Postby Alice Chang 2H » Sun Jan 12, 2020 4:27 pm

When a reaction is at equilibrium, both sides of the equation (reactants and products) will be reacting, but at the same rate (thus being equal). If you find that K is equal to 1, then the reaction is at equilibrium.

Solving for Q means solving for the ratio between products and reaction at any time of the reaction. If Q=K, then you are at equilibrium.

I think it's just a matter of knowing K, and figuring out if Q=K or not. I don't think any questions would be too tricky in giving vague information about the reaction.

Hussain Chharawalla 1G
Posts: 100
Joined: Sat Jul 20, 2019 12:15 am

Re: Equilibrium vs. Non-Equilibrium

Postby Hussain Chharawalla 1G » Sun Jan 12, 2020 4:32 pm

To add on, if Q<K that means there is either too much reactants or not enough products. So the reaction will continue in the forward direction.

If Q>K, then there is too many products and not enough reactants, so the reverse reaction is favored.

Emil Velasco 1H
Posts: 96
Joined: Wed Nov 21, 2018 12:19 am

Re: Equilibrium vs. Non-Equilibrium

Postby Emil Velasco 1H » Sun Jan 12, 2020 4:47 pm

If Q does not equal K or vice versa, then I believe you would have to solve for either respectively. If Q is the same value calculated, then the reaction is at equilibrium


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