At $300 \mathrm{~K}, \mathrm{~K}_{\mathrm{C}}$ for the reaction. $\mathrm{A}_2 \mathrm{~B}_2(\mathrm{~g})…

At $300 \mathrm{~K}, \mathrm{~K}_{\mathrm{C}}$ for the reaction. $\mathrm{A}_2 \mathrm{~B}_2(\mathrm{~g}) \rightleftharpoons \mathrm{A}_2(\mathrm{~g})+\mathrm{B}_2(\mathrm{~g})$ is $100 \mathrm{~mol} \mathrm{~L}^{-1}$. What is its $\mathrm{K}_{\mathrm{p}}$ (in atm) at the same temperature? $\left(\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm}^{-1} \mathrm{~mol}^{-1}\right)$
  1. 100
  2. 2460
  3. 4.06
  4. 246

Solution

$\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} \times(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g}}}$ Given, $\begin{aligned} & \mathrm{K}_{\mathrm{c}}=100 \mathrm{~mol} \mathrm{~L}^{-1} \\ & \mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \\ & \mathrm{~T}=300 \mathrm{~K} \\ & \Delta \mathrm{n}=(\text { no. of gaseous product })-(\text { no. of gaseous reactant }) \\ & =2-1=1 \\ & \therefore \quad \mathrm{~K}_{\mathrm{P}}=100 \mathrm{~mol} \mathrm{~L}^{-1} \times\left(0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 300 \mathrm{~K}\right)^1 \\ & \mathrm{~K}_{\mathrm{P}}=2460 \mathrm{~atm} \end{aligned}$

Asked in: AP EAMCET 2024 (23 May Shift 1)

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