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JEE Main 20245 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
5 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Main 2024 · 5 April · Evening Shift · Chemistry · Chemical Kinetics

Consider the following single step reaction in gas phase at...

Consider the following single step reaction in gas phase at constant temperature. (2  A(g)+B(g)→C(g)\mathrm{~A}_{(\mathrm{g})}+\mathrm{B}_{(\mathrm{g})} \rightarrow \mathrm{C}_{(\mathrm{g})}) The initial rate of the reaction is recorded as (\mathrm{r}}_{1}) when the reaction starts with (1.5 \mathrm{~atm}) pressure of (\mathrm{A})and(0.7 atm)pressureof() and (0.7 \mathrm{~atm}) pressure of (\mathrm{B}).Aftersometime,therate(r). After some time, the rate (\mathrm{r}}_{2}) is recorded when the pressure of (C\mathrm{C}) becomes (0.5  atm\mathrm{~atm}). The ratio (\mathrm{r}}_{1}: \mathrm{r}}_{2}) is ______ (×\times10^{-1}). (Nearest integer)

Correct answer

0

Step-by-step solution

2 A( g)+B(g)⟶C(g)2 \mathrm{~A}(\mathrm{~g})+\mathrm{B}(\mathrm{g}) \longrightarrow \mathrm{C}(\mathrm{g})
r11.5 atm0.7 atmr20.5 atm0.2 atm0.5 atm∵r=K[PA]2[PB]r1=K[1.5]2[0.7]r2=K[0.5]2[0.2]r1r2=9×72=31.5=315×10−1\begin{array}{ll} \mathrm{r}_1 & 1.5 \mathrm{~atm} \quad 0.7 \mathrm{~atm} \\ \mathrm{r}_2 & 0.5 \mathrm{~atm} \quad 0.2 \mathrm{~atm} \quad 0.5 \mathrm{~atm} \\ & \because \mathrm{r}=\mathrm{K}\left[\mathrm{P}_{\mathrm{A}}\right]^2\left[\mathrm{P}_{\mathrm{B}}\right] \\ & \mathrm{r}_1=\mathrm{K}[1.5]^2[0.7] \\ & \mathrm{r}_2=\mathrm{K}[0.5]^2[0.2] \\ & \frac{\mathrm{r}_1}{\mathrm{r}_2}=9 \times \frac{7}{2}=31.5=315 \times 10^{-1} \end{array} Ans. 315
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