At equilibrium,the concentrations are [N_2] = | vedclass

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(B) Given,$[N_2] = 3.0 	imes 10^{-3} \ M$,$[O_2] = 4.2 	imes 10^{-3} \ M$,and $[NO] = 2.8 	imes 10^{-3} \ M$.For the reaction $N_{2(g)} + O_{2(g)}

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$0.622$

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(B) Given,$[N_2] = 3.0 	imes 10^{-3} \ M$,$[O_2] = 4.2 	imes 10^{-3} \ M$,and $[NO] = 2.8 	imes 10^{-3} \ M$.For the reaction $N_{2(g)} + O_{2(g)} ightleftharpoons 2NO_{(g)}$,the equilibrium constant $K_C$ is given by:$K_C = \frac{[NO]^2}{[N_2][O_2]}$Substituting the values:$K_C = \frac{(2.8 	imes 10^{-3})^2}{(3.0 	imes 10^{-3})(4.2 	imes 10^{-3})} = \frac{7.84 	imes 10^{-6}}{12.6 	imes 10^{-6}} \approx 0.622$The relationship between $K_p$ and $K_C$ is $K_p = K_C(RT)^{\Delta n}$.Here,$\Delta n = (	ext{moles of gaseous products}) - (	ext{moles of gaseous reactants}) = 2 - (1 + 1) = 0$.Since $\Delta n = 0$,$K_p = K_C(RT)^0 = K_C$.Therefore,$K_p = 0.622$.

At equilibrium,the concentrations are $[N_2] = 3.0 \times 10^{-3} \ M$,$[O_2] = 4.2 \times 10^{-3} \ M$,and $[NO] = 2.8 \times 10^{-3} \ M$ in a sealed vessel at $800 \ K$ and $1 \ atm$ pressure. What will be $K_p$ for the given reaction?
$N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$

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At equilibrium,the concentrations are $[N_2] = 3.0 \times 10^{-3} \ M$,$[O_2] = 4.2 \times 10^{-3} \ M$,and $[NO] = 2.8 \times 10^{-3} \ M$ in a sealed vessel at $800 \ K$ and $1 \ atm$ pressure. What will be $K_p$ for the given reaction?$N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$