JEE Main 2024 6 April Evening Shift
Exam JEE Main
Year 2024
Date 6 April
Shift Evening Shift
Subject Chemistry
Chapter Chemical Equilibrium JEE Main 2024 · 6 April · Evening Shift · Chemistry · Chemical Equilibrium
The ratio ((K_P)/(K_C)) for the reaction : (CO_ (g) +(1)/(2) O_...
The ratio (\frac{K_P}{K_C}) for the reaction : (\mathrm{CO}_{(\mathrm{g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{~g})}) is :
Options A (\frac{1}{\sqrt{\mathrm{RT}}}) B ((\mathrm{RT})^{1 / 2}) C (\mathrm{RT}) D 1 Correct answer A. (\frac{1}{\sqrt{\mathrm{RT}}})
Step-by-step solution
C O ( g ) + 1 2 O 2 ( g ) ⇌ C O 2 ( g ) Δ n g = 1 − ( 1 + 1 2 ) = − 1 2 K p K C = ( R T ) Δ n g = 1 R T \begin{aligned} & \mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{~g}) \\ & \Delta \mathrm{n}_{\mathrm{g}}=1-\left(1+\frac{1}{2}\right)=-\frac{1}{2} \\ & \frac{\mathrm{K}_{\mathrm{p}}}{\mathrm{K}_{\mathrm{C}}}=(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g}}}=\frac{1}{\sqrt{\mathrm{RT}}}\end{aligned} CO ( g ) + 2 1 O 2 ( g ) ⇌ CO 2 ( g ) Δ n g = 1 − ( 1 + 2 1 ) = − 2 1 K C K p = ( RT ) Δ n g = RT 1
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