JEE MAIN - Chemistry (2024 - 9th April Morning Shift - No. 26)
Given below are two statements :
Statement I : The rate law for the reaction $$A+B \rightarrow C$$ is rate $$(r)=k[A]^2[B]$$. When the concentration of both $$\mathrm{A}$$ and $$\mathrm{B}$$ is doubled, the reaction rate is increased "$$x$$" times.
Statement II : _9th_April_Morning_Shift_en_26_1.png)
The figure is showing "the variation in concentration against time plot" for a "$$y$$" order reaction.
The Value of $$x+y$$ is __________.Explanation
$$\begin{aligned} & x=8, y=0 \\ & \frac{r_2}{r_1}=\left(\frac{C_{\mathrm{A}_2}}{C_{\mathrm{A}_1}}\right)^2\left(\frac{\mathrm{C}_{\mathrm{B}_2}}{\mathrm{C}_{\mathrm{B}_1}}\right) \\ & \frac{\mathrm{r}_2}{\mathrm{r}_1}=\left(\frac{2 \mathrm{C}_{\mathrm{A}_1}}{\mathrm{C}_{\mathrm{A}_1}}\right)^2\left(\frac{2 \mathrm{C}_{\mathrm{B}_1}}{\mathrm{C}_{\mathrm{B}_1}}\right) \\ & \mathrm{r}_2=8 \mathrm{r}_1 \\ & x=8 \end{aligned}$$
_9th_April_Morning_Shift_en_26_2.png)
$$\begin{aligned} & \Rightarrow \text { Zero order } \\ & y=0 \end{aligned}$$
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