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AURORKA [14]
4 years ago
6

A reversible reaction has a forward rate constant of 0.412 mol/L/s and a reverse reaction rate constant of 0.827 mol/L/s. What's

the equilibrium constant for this reaction?
Chemistry
1 answer:
andrey2020 [161]4 years ago
7 0

Answer: 0.498

Explanation:

Given the following :

Forward rate constant (Kf) = 0.412 mol/L/s

Reverse rate constant (Kr) = 0.827 mol/L/s

The ratio of the rate constant for the forward reaction and the rate constant for the reverse reaction produces the equilibrium constant.

Therefore,

Keq = (forward rate constant (Kf) / Reverse rate constant)

Keq = 0.412mol/L/s / 0.827mol/L/s

Keq = 0.498

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