We are asked what is the new equilibrium equation after multiplying the reaction by a factor of 2.

2A + B ⇌ 4C

multiply by 2:

(2A + B ⇌ 4C) x 2

**4A + 2B ⇌ 8C**

Get the K_{c} expression:

$\overline{){{\mathbf{K}}}_{{\mathbf{c}}}{\mathbf{}}{\mathbf{=}}\frac{\mathbf{products}}{\mathbf{reactants}}}$

**The new Kc expression is:**

${\mathbf{K}}_{\mathbf{c}}\mathbf{}\mathbf{=}\frac{{\mathbf{\left[}\mathbf{C}\mathbf{\right]}}^{\mathbf{8}}}{{\mathbf{\left[}\mathbf{A}\mathbf{\right]}}^{\mathbf{4}}{\mathbf{\left[}\mathbf{B}\mathbf{\right]}}^{\mathbf{2}}}$

Consider the equilibrium reaction

2A + B ⇌ 4C

After multiplying the reaction by a factor of 2, what is the new equilibrium equation?

Drag the following labels into the numerator and denominator as needed to create the equilibrium-constant expression, K_{c}, for the new equilibrium reaction,

If the initial reaction contains 1.31 M A, 1.21 M B, and 2.51 M C, calculate K0 for the new equilibrium reaction.

K_{c} =

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