The Kc for the reaction N2O4(g)⇌2NO2(g) is 4.62×10−3M at 298 K. Its KP value would be

# The ${\mathrm{K}}_{\mathrm{c}}$ for the reaction ${\mathrm{N}}_{2}{\mathrm{O}}_{4}\left(\mathrm{g}\right)⇌2{\mathrm{NO}}_{2}\left(\mathrm{g}\right)$ is $4.62×{10}^{-3}\mathrm{M}$ at 298 K. Its ${\mathrm{K}}_{\mathrm{P}}$ value would be

1. A

11.45 kPa

2. B

$1.86×{10}^{-6}\mathrm{kPa}$

3. C

$0.07{\mathrm{kPa}}^{-1}$

4. D

0.038 kPa

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### Solution:

${K}_{p}={K}_{c}\left(RT{\right)}^{\mathrm{\Delta }{\nu }_{\mathrm{E}};}$ where $\mathrm{\Delta }{\nu }_{\mathrm{g}}=+1$

$=\left(4.62×{10}^{-3}{\mathrm{molL}}^{-1}\right)\left(8.314{\mathrm{kPaLK}}^{-1}{\mathrm{mol}}^{-1}\right)\left(298\mathrm{K}\right)=11.45\mathrm{kPa}$

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