Consider the graph given in the following dia | vedclass

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(D) The instantaneous rate of reaction at a specific time is determined by the slope of the tangent drawn to the concentration-time curve at that poin

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$\frac{C_4 - C_2}{50 - 30}$

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(D) The instantaneous rate of reaction at a specific time is determined by the slope of the tangent drawn to the concentration-time curve at that point.For the $40^{th}$ second,the tangent passes through points corresponding to $(30, C_2)$ and $(50, C_5)$.The slope of this tangent is calculated as $\frac{\Delta C}{\Delta t} = \frac{C_5 - C_2}{50 - 30}$.Option $(A)$ represents this slope correctly.Options $(B)$ and $(C)$ represent slopes of secant lines that approximate the instantaneous rate at $40 \ s$ by using points on either side of $40 \ s$ (e.g.,$(30, C_2)$ to $(40, C_3)$ or $(20, C_1)$ to $(40, C_3)$).Option $(D)$,$\frac{C_4 - C_2}{50 - 30}$,uses $C_4$ instead of $C_5$,which does not correspond to the tangent line at $40 \ s$. Therefore,it does not represent the instantaneous rate.

Consider the graph given in the following diagram. Which of the following options does not show the instantaneous rate of reaction at the $40^{th}$ second?

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