Calculate Kc with dissociation degree If I have the reaction
$$\ce{2HI (g) ⇌ H2(g) + I2(g)}$$
and know that at $T = \pu{448 °C}$ the dissociation degree $α = 0.2198,$ how do I calculate $K_c?$
I thought that it would be something like this:
$$
\begin{array}{cccCc}
\ce{&amp;2HI &amp;&lt;=&gt; &amp;H2(g) &amp;+ &amp;I2(g)} \\
&amp;2n &amp;&amp; 0 &amp;&amp; 0 \\
&amp;2n(1 - α) &amp;&amp; αn &amp;&amp; αn \\
\end{array}
$$
Then
$$n_\mathrm{tot} = 2n(1 - α) + 2nα = 2n$$
$$p(\ce{HI}) = \frac{2nα}{2n} = (1 - α)p_\mathrm{tot}$$
$$p\ce{I2} = p(\ce{H2}) = αp_\mathrm{tot}$$
$$K_p = \frac{(αp_\mathrm{tot})^2}{((1 - α)p_\mathrm{tot})^2} = \frac{α^2}{(1-α)^2} = \frac{0.2198^2}{(1 - 0.2198)^2} = 0.0794$$
and then from $K_p$ I could calculate $K_c:$
$$K_c = \frac{0.0794}{(8.314 \times 721)^0}$$
I think the last step is where I'm going wrong since it shouldn't be raised to zero. The correct answer is $K_c = 0.01984.$
You don't need gas laws, temperature or $K_p.$
You overcomplicated the problem: $K_c$ can be found by following its definition with an RICE table:
$$
\begin{array}{lccccc}
&amp;\text{R}\ce{&amp;2 HI(g) &amp;&lt;=&gt; &amp;H2(g) &amp;+ &amp;I2(g)} \\
&amp;\text{I} &amp; 2c_0 &amp;&amp; 0 &amp;&amp; 0 \\
&amp;\text{C} &amp; -2\alpha c_0 &amp;&amp; \alpha c_0 &amp;&amp; \alpha c_0 \\
&amp;\text{E} &amp; 2(1 - \alpha)c_0 &amp;&amp; \alpha c_0 &amp;&amp; \alpha c_0 \\
\end{array}
$$
For the sake of consistency, I denoted the initial concentration with $c_0$ and dissociation degree as $\alpha.$
By definition, the equilibrium constant on a concentration basis $K_c$ can be found as follows:
$$K_c = \frac{[\ce{H2(g)}][\ce{I2(g)}]}{[\ce{HI(g)}]^2} = \frac{(\alpha c_0)^2}{(2(1 - \alpha)c_0)^2} = \frac{\alpha^2}{4(1 - \alpha)^2} = \frac{0.2198^2}{4(1 - 0.2198)^2} \approx 0.0198$$