A graph plotted between log t50% vs log of concentration is a straight line. What conclusion can you draw from the given graph?

Engineering

Chemistry

Order of Reaction and Law of Mass Action

Methods of Determination of Order of Reaction

Calculation of First Order Rate Constant using Different Parameters

Question

A graph plotted between log t_50% vs log of concentration is a straight line. What conclusion can you draw from the given graph?

A

$n = 1, t_{1 / 2} = \frac{1}{k - a}$

B

n = 2, t_1/2 = 1/a

C

n = 2, t_1/2 = 0.639/k

D

none of these

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For an nth order reaction, the half-life (t_1/2) is related to the initial concentration (a) by the equation: $t_{1 / 2} \propto \frac{1}{a^{n - 1}}$ . Taking the logarithm of both sides gives: $\log t_{1 / 2} = \log constant + (1 - n) \log a$ . A plot of log t_1/2 vs log a is a straight line with slope = (1-n). The graph shows a negative slope, indicating n>1. The specific option with n=2 and t_1/2 = 1/(k a) (or 0.693/k for first order) is incorrect. The correct relationship for a second-order reaction is t_1/2 = 1/(k a).

Final Answer: none of these