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Anonymous asked in Science & MathematicsChemistry · 2 months ago

Catalyst and equilibrium?

For a different reaction, Kc = 112, kf=6.56×103s^−1 , and kr= 58.5 s^−1 . Adding a catalyst increases the forward rate constant to 4.07×105 s−1 . What is the new value of the reverse reaction constant, kr, after adding catalyst?

It is also asking for the units for kr... but I'm assuming it's going to be s^-1 since they should cancel out when getting the K constant.

1 Answer

  • 2 months ago
    Favorite Answer

    The catalyst of a reaction increases the rates of the forward reaction and the reverse reaction to the same extent, and does not affect the Kc at constant temperature.

    Kc = kf/kr

    Then, kr = kf/Kc

    New reverse reaction rate constant. kr = (4.07 × 10⁵)/112 s⁻¹ = 3.63 × 10³ s⁻¹

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