Problem: Consider the reaction: A(g) ⇌ 2B(g). The graph plots the concentrations of A and B as a function of time at a constant temperature.What is the equilibrium constant for this reaction at this temperature?

FREE Expert Solution

At equilibrium

  • rate of forward reaction = rate of reverse reaction
  • no observable change in concentration of products/reactants
    • [A] = 0.6M
    • [B] = 0.775 M


  • Forward: A(g) → 2B(g)
  • Reverse: 2B(g) A(g) 
    • rate law = k[reactants]


ratefwd=kfwd[A]     raterev=krev[B]2


For a one step reaction, equilibrium constant K is: 

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Problem Details
Consider the reaction: A(g) ⇌ 2B(g). The graph plots the concentrations of A and B as a function of time at a constant temperature.
The figure shows the concentration as a function of time. Concentration is measured from 0 to 1 mole per liter on the y-axis and time is measured on the x-axis. There are two curves on the plot. Curve A starts at 1 mole per liter, then decreases to approximately 0.6 moles per liter, and then becomes a straight horizontal line. Curve B starts at 0 moles per liter, increases with the same speed as curve A to 0.8 moles per liter, and then becomes a straight horizontal line. These curves cross at the concentration about 0.65 moles per liter.

What is the equilibrium constant for this reaction at this temperature?

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