37.8 mathrm~g mathrm~N_2 mathrmO_5$37.8 \mathrm{~g} \mathrm{~N}_{2} \mathrm{O}_{5}$ was taken in a 1 mathrm~L$1 \mathrm{~L}$ reaction vessel and allowed to undergo the following reaction at 500 mathrm~K$500 \mathrm{~K}$: 2 mathrm N _ 2 mathrm O _ 5 (mathrm g) ightarrow 2 mathrm N _ 2 mathrm O _ 4 (mathrm g) + mathrm O _ 2 (mathrm g)$$2 \mathrm {N} _ {2} \mathrm {O} _ {5 (\mathrm {g})} ightarrow 2 \mathrm {N} _ {2} \mathrm {O} _ {4 (\mathrm {g})} + \mathrm {O} _ {2 (\mathrm {g})}$$ The total pressure at equilibrium was found to be 18.65 bar. Then, mathrmKp = \_ \_ \_ \_ times 10^-2$\mathrm{Kp} = \_ \_ \_ \_ \times 10^{-2}$ [nearest integer] Assume mathrmN_2mathrmO_5$\mathrm{N}_2\mathrm{O}_5$ to behave ideally under these conditions Given: mathrmR = 0.082$\mathrm{R} = 0.082$ bar mathrmL \, mathrmmol^-1 \, mathrmK^-1$\mathrm{L} \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1}$

### Related Formula K_C = frac[textProducts][textReactants]$$K_C = \frac{[\text{Products}]}{[\text{Reactants}]}$$ ### Core Logic K_C = fractextProducts ion conc.textReactants ion conc.$$K_C = \frac{\text{Products ion conc.}}{\text{Reactants ion conc.}}$$ K_C = frac[FeSCN^2+][Fe^3+][SCN^-]$$K_C = \frac{[FeSCN^{2+}]}{[Fe^{3+}][SCN^{-}]}$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium