18 July, 09:05
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?
18 July, 11:00
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.
Keq = (forward rate constant (Kf) / Reverse rate constant)
Keq = 0.412mol/L/s / 0.827mol/L/s
Keq = 0.498
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» 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?