Q12 / 160 · AP EAMCET 2023 (18 May Shift 1) · Chemistry

At $1000 \mathrm{~K}$, the value of $\mathrm{K}_{\mathrm{c}}$ for the below reaction is 10 $\mathrm{mol} \mathrm{L}^{-1}$. Value of $\mathrm{K}_{\mathrm{p}}$ (in atm) is $\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})$ (given $\mathrm{R}=0.082 \mathrm{~atm} \mathrm{~L} \mathrm{~mol}{ }^{-1} \mathrm{~K}^{-1}$ )

  • A
    $82$
  • B
    $0.82$
  • C
    $8.2$
  • Correct answer
    D
    $820$

Solution

$K_p=K_c(R T)^{\Delta n_g}$ $\begin{aligned} & \mathrm{K}_{\mathrm{c}}=10 \mathrm{~mol} \mathrm{~L}^{-1}, \mathrm{~T}=1000 \mathrm{~K} \\ & \mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \\ & \Delta \mathrm{n}_{\mathrm{g}}=(1+1)-1=1\end{aligned}$ $\Rightarrow K_p=(10)(0.082 \times 1000)^1=820$

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

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