(B) The relationship between the equilibrium constant $(K_c)$,forward rate constant $(K_f)$,and backward rate constant $(K_b)$ is given by the express

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(B) The relationship between the equilibrium constant $(K_c)$,forward rate constant $(K_f)$,and backward rate constant $(K_b)$ is given by the expression: $K_c = \frac{K_f}{K_b}$.Given that $K_c = 100$ and $K_f = 10^5$,we can rearrange the formula to solve for $K_b$:$K_b = \frac{K_f}{K_c} = \frac{10^5}{100} = \frac{10^5}{10^2} = 10^3$.Therefore,the rate constant for the backward reaction is $10^3$.

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Kp and Kc Relationship

Medium

The equilibrium constant $K_c$ for the reaction $HA + B \rightleftharpoons BH^{+} + A^{-}$ is $100$. If the rate constant for the forward reaction is $10^5$,then the rate constant for the backward reaction is:

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A
$10^7$
B
$10^3$
C
$10^{-3}$
D
$10^{-5}$

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6-1.Equilibrium (Chemical Equilibrium)

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Kp and Kc Relationship

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The equilibrium constant $K_c$ for the reaction $HA + B \rightleftharpoons BH^{+} + A^{-}$ is $100$. If the rate constant for the forward reaction is $10^5$,then the rate constant for the backward reaction is:

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