A) \[2\times {{10}^{-4}}\]
B) \[2.80\times {{10}^{-6}}\]
C) \[2.80\times {{10}^{-4}}\]
D) \[2.\times {{10}^{-6}}\]
Correct Answer: A
Solution :
| [a] Applying the relationship |
| \[{{K}_{P}}={{K}_{C}}{{\left( RT \right)}^{\Delta n}}g\] |
| \[{{\text{K}}_{\text{C}}}\text{=0}\text{.50;R=0}\text{.082 L atm }{{\text{K}}^{\text{-1}}}\text{mo}{{\text{l}}^{\text{-1}}}\] |
| \[T=\left( 400+273 \right)=673;\Delta ng=\left( 2-4 \right)=-2\] |
| \[\therefore \] \[{{K}_{P}}=0.5{{\left( 0.082\times 673 \right)}^{-2}}\] |
| \[=0.5{{\left( 55.186 \right)}^{-2}}=2\times {{10}^{-4}}\] |
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