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]**

$\overline{){{\mathbf{rate}}}_{{\mathbf{fwd}}}{\mathbf{=}}{{\mathbf{k}}}_{{\mathbf{fwd}}}\mathbf{\left[}\mathbf{A}\mathbf{\right]}}$ $\overline{){{\mathbf{rate}}}_{{\mathbf{rev}}}{\mathbf{=}}{{\mathbf{k}}}_{{\mathbf{rev}}}{\mathbf{\left[}\mathbf{B}\mathbf{\right]}}^{\mathbf{2}}}$

For a one step reaction,** equilibrium constant K** is:

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?

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