The rate of a certain reaction depends on con | vedclass

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(A) The given rate equation is $-dc/dt = (K_1 C) / (1 + K_2 C)$.When the concentration $(C)$ is very high,the term $(K_2 C)$ becomes much larger than

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zero order

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(A) The given rate equation is $-dc/dt = (K_1 C) / (1 + K_2 C)$.When the concentration $(C)$ is very high,the term $(K_2 C)$ becomes much larger than $1$,i.e.,$(K_2 C >> 1)$.Therefore,the denominator $(1 + K_2 C)$ can be approximated as $(K_2 C)$.Substituting this into the rate equation: $-dc/dt \approx (K_1 C) / (K_2 C) = K_1 / K_2$.Since the rate is now independent of the concentration $(C)$,the reaction follows zero order kinetics.

The rate of a certain reaction depends on concentration according to the equation $-dc/dt = (K_1 C) / (1 + K_2 C)$. What is the order when concentration $(C)$ is very high?

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