Q.75. For a reaction A P at T K, the half life (t) is plotted as a function of initial concentration [A] of A as given below : [Figure: Graph showing half-life (t) on y-axis and initial concentration [A] on x-axis. The graph is a straight line passing through (1.5x10^-3, 240) and (4x10^-3, x).]

Sol. From the plot, $t_{1/2} \propto [A]_0^{1-N}$ $1-N=1$ $N = 0$ (order) $t_{1/2} = \frac{[A]_0}{2K}$ $240 = \frac{4 \times 10^{-3}}{2K}$ ....(1) $x = \frac{1.5 \times 10^{-3}}{2K}$ ....(2) Divide (1) / (2) $\frac{240}{x} = \frac{4}{1.5}$ $x = 90$

What is the correct answer for JEE Main Chemistry Q75?

The correct answer is 90. Sol. From the plot, $t_{1/2} \propto [A]_0^{1-N}$ $1-N=1$ $N = 0$ (order) $t_{1/2} = \frac{[A]_0}{2K}$ $240 = \frac{4 \times 10^{-3}}{2K}$ ....(1) $x = \frac{1.5 \times 10^{-3}}{2K}$ ....(2) Divide (1) / (2) $\frac{240}{x} = \frac{4}{1.5}$ $x = 90$

Q.75. For a reaction A P at T K, the half life (t) is plotted as a function of initial concentration [A] of A as given below : [Figure: Graph showing half-life (t) on y-axis and initial…

The correct answer is 90. Sol. From the plot, $t_{1/2} \propto [A]_0^{1-N}$ $1-N=1$ $N = 0$ (order) $t_{1/2} = \frac{[A]_0}{2K}$ $240 = \frac{4 \times 10^{-3}}{2K}$ ....(1) $x = \frac{1.5 \times 10^{-3}}{2K}$ ....(2) Divide (1) / (2) $\frac{240}{x} = \frac{4}{1.5}$ $x = 90$

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Q.75. For a reaction A $\rightarrow$ P at T K, the half life (t$_{1/2}$) is plotted as a function of initial concentration [A]$_0$ of A as given below : [Figure: Graph showing half-life (t$_{1/2}$) on y-axis and initial concentration [A]$_0$ on x-axis. The graph is a straight line passing through (1.5x10^-3, 240) and (4x10^-3, x).]

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