When Q>K, is the rate of the forward reaction greater than the rate of the reverse reaction?

asked by @raadn1 • 9 months ago • Chemistry • 5 pts
  • Note: It's for equilibrium Raad commented 9 months ago
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2 answers

Hey Raad! So remember that Q represents the reaction quotient while K represents the equilibrium constant. Q is calculated using the initial values (whether it is concentration or pressure Qc or Qp, respectively). K is calculated using the values at equilibrium whether it is Kc or Kp.

In this problem, Q is said to be greater than K. That means that there are more products than reactants. Remember that:

Q= products / reactants

So if Q is very large, it means we must have many products and little to no reactants.

Notice if you were to draw a number line:

____K____Q

so the arrow would go toward the left in order to achieve equilibrium. It also looks a lot like K < Q (it's a good trick to remember that the reaction proceeds left!):

____K<-----Q

This means that the reaction will move left to make more reactants since we initially have more products.

Q>K favors the reverse reaction. This means that the rate of the reverse reaction will be greater than the rate of the forward reaction.

Hope this helps! Let us know if you have any other questions :)

answered by @sabrina • 9 months ago

Hey there Raad!

If the Reaction quotient (Q) is larger than equilibrium constant (K), than the reaction will be shifting to the left to restore equilibrium. This means that until equilibrium is reached, the rate of the reverse reaction will be greater than the forward.

Hope this helps!

answered by @dashab2 • 9 months ago