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Thread: How to solve the problem of Relative Concentrations at Equilibrium?

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    Level 7 - I know you and your Friends michele's Avatar
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    How to solve the problem of Relative Concentrations at Equilibrium?

    Reactant A undergoes dissociation into products B and C according to the following equilibrium reaction equation

    A(g)=B(g) + C(g)
    A rigid vessel, initially containing 0.042 mol/L of pure A, is allowed to equilibrate at 72°C.

    What would be the value of the equilibrium constant (Keq expressed in terms of the molar concentrations) if the equilibrium concentrations of all species are equal.
    What would be the value of the equilibrium constant (Keq expressed in terms of the molar concentrations) if, at equilibrium, the concentration of A is 1.0% that of C?

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    Level 15 - A Legend cncp's Avatar
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    If eqm. concs are equal, 0.021 mol of A remain and 0.021 mol of each of B and C have been created so K = [B].[C]/[A] = 0.021*0.021/0.021 = 0.021

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