The values of

Question:

The values of $\mathrm{K}_{\mathrm{p}} / \mathrm{K}_{\mathrm{c}}$ for the following reactions at 300 $\mathrm{K}$ are, respectively: (At $300 \mathrm{~K}, \mathrm{RT}=24.62 \mathrm{dm}^{3} \mathrm{~atm}$ $\mathrm{mol}^{-1}$ )

  1. $1,24.62 \mathrm{dm}^{3} \mathrm{~atm} \mathrm{~mol}^{-1}$$606.0 \mathrm{dm}^{6} \mathrm{~atm}^{2} \mathrm{~mol}^{-2}$

  2. $1,24.62 \mathrm{dm}^{3} \mathrm{~atm} \mathrm{~mol}^{-1}$,$1.65 \times 10^{-3} \mathrm{dm}^{-6} \mathrm{~atm}^{-2} \mathrm{~mol}^{2}$

  3. $1,4.1 \times 10^{-2} \mathrm{dm}^{-3} \mathrm{~atm}^{-1} \mathrm{~mol}$$606 \mathrm{dm}^{6} \mathrm{~atm}^{2} \mathrm{~mol}^{-2}$

  4. $24.62 \mathrm{dm}^{3} \mathrm{~atm} \mathrm{~mol}^{-1}$,$606.0 \mathrm{dm}^{6} \mathrm{~atm}^{2} \mathrm{~mol}^{-2}$,$1.65 \times 10^{-3} \mathrm{dm}^{-6} \mathrm{~atm}^{-2} \mathrm{~mol}^{2}$


Correct Option: , 2

Solution:

$\mathrm{K}_{\mathrm{p}}=K_{c}(\mathrm{RT})^{\Delta n_{g}}$

$\Delta \mathrm{n}_{\mathrm{g}}=$ No. of gaseous moles of products

- No. of gaseous moles of reactants

$\frac{K_{p}}{K_{c}}=(\mathrm{RT})^{\Delta^{n} g}$

$\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}), \Delta n_{g}=0$

$=\left(24.62 \mathrm{dm}^{3} \mathrm{~atm} \mathrm{~mol}^{-1}\right)^{0}=1$

$\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}), \Delta n_{g}=1$

$\frac{K_{p}}{K_{c}}=24.62 \mathrm{dm}^{3} \mathrm{~atm} \mathrm{~mol}^{-1}$

$\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}), \Delta n_{g}=-2$

$\frac{K_{p}}{K_{c}}=\left(24.62 \mathrm{dm}^{-6} \mathrm{~atm}^{-2} \mathrm{~mol}^{-2}\right)^{-2}$

$=\frac{1}{\left(24.62 \mathrm{dm}^{2} \mathrm{~atm} \mathrm{~mol}^{-1}\right)^{2}}$

$=1.65 \times 10^{-3} \mathrm{dm}^{-6} \mathrm{~atm}^{-2} \mathrm{~mol}^{-2}$

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