At 300 ~K, ~K_C for the reaction. A_2 ~B_2(~g) leftharpoons A_2(~g)+B_2(~g) is 100 ~mol ~L^-1 . What is its K_p (in atm) at the same temperature? (R=0.082 ~L ~atm^-1 ~mol^-1 )

Correct answer: (B) 2460. K_p=K_c ×(RT)^Δ n_g Given, K_c=100 ~mol ~L^-1 R=0.082 ~L ~atm ~mol^-1 ~K^-1 ~T=300 ~K Δ n=( no. of gaseous product )-( no. of gaseous reactant ) =2-1=1 ∴ ~K_P=100 ~mol ~L^-1 × (0.082 ~L ~atm ~mol^-1 ~K^-1 × 300 ~K )^1 ~K_P=2460 ~atm

Q11 / 160 · AP EAMCET 2024 (23 May Shift 1) · Chemistry

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? $\left(\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm}^{-1} \mathrm{~mol}^{-1}\right)$

A
100
Correct answer
B
2460
C
4.06
D
246

Solution

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

Hence, the correct answer is option: 'B'.

Solution

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