$N_{2}=3.0 \times 10^{-3} M$
$O_{2}=4.2 \times 10^{-3} M$
and $N O=2.8 \times 10^{-3} M$
in a sealed vessel at $800 \,K$ and $1$ $atm$ pressure.........$atm$ will be $K_{p}$ for the given reaction?
$N_{2}(g)+O_{2}(g) \rightleftharpoons 2 N O(g)$
$O_{2}=4.2 \times 10^{-3} M$
and $N O=2.8 \times 10^{-3} M$
For the given reaction,
$N_{2}(g)+O_{2}(g) \rightleftharpoons 2 N O(g)$
equilibrium constant $K_{C}$ can be written as
$K_{C}=\frac{[N O]^{2}}{\left[N_{2}\right]\left[O_{2}\right]}$
$\therefore K_{C}=\frac{\left(2.8 \times 10^{-3} M\right)^{2}}{\left(3.0 \times 10^{-3} M\right)\left(4.2 \times 10^{-3} M\right)}=0.622$
$\because K_{p}=K_{C} \cdot(R T)^{\Delta n}$
$\Delta n=$ Number of moles of gaseous products number of moles of gaseous reactants
$\Delta n=2-2=0$
$\therefore K_{p}=K_{C^{.}}(R T)^{o}$
$K_{p}=K_{C}$ or, $K_{p}=0.622$ $atm$
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| $(IE_1 + IE_2)$ | $(IE_3 + IE_4)$ |
| $(P)$ $2.45\, KJ/mol$ | $8.82\, KJ/mol$ |
| $(Q)$ $2.85\, KJ/mol$ | $6.11\, KJ/mol$ |
then according to given information the incorrect statements is/are
