Question
Determine the mole fraction of sodium that ionizes according to the reaction $\mathrm{Na} \rightleftharpoons \mathrm{Na}^{+}+e^{-}$ at $2000 \mathrm{K}$ and $0.8 \operatorname{atm}\left(K_{P}=0.668 \text { for this reaction }\right)$.
Step 1
Let's denote the initial amount of $\mathrm{Na}$ as 'a', and the amount that ionizes as 'b'. So, the amount of $\mathrm{Na}$ left after ionization is $a-b$ and the amounts of $\mathrm{Na}^{+}$ and $e^{-}$ formed are 'b'. Show more…
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