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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:

$$ \begin{array}{cccCc} \ce{&2HI &<=> &H2(g) &+ &I2(g)} \\ &2n && 0 && 0 \\ &2n(1 - α) && αn && αn \\ \end{array} $$

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 $K_c = 0.01984.$

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  • $\begingroup$ At $\alpha=1$, i.e. at full conversion, you errorneously suppose $p_{\ce{H2}}=p_\mathrm{tot}$ and $p_{\ce{I2}}=p_\mathrm{tot}$. Should be $p_{\ce{I2}} = p_{\mathrm{H2}} = \frac{\alpha}{2} * p_\mathrm{tot}$ $\endgroup$
    – Poutnik
    Jan 28, 2021 at 15:48

1 Answer 1

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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:

$$ \begin{array}{lccccc} &\text{R}\ce{&2 HI(g) &<=> &H2(g) &+ &I2(g)} \\ &\text{I} & 2c_0 && 0 && 0 \\ &\text{C} & -2\alpha c_0 && \alpha c_0 && \alpha c_0 \\ &\text{E} & 2(1 - \alpha)c_0 && \alpha c_0 && \alpha c_0 \\ \end{array} $$

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:

$$K_c = \frac{[\ce{H2(g)}][\ce{I2(g)}]}{[\ce{HI(g)}]^2} = \frac{(\alpha c_0)^2}{(2(1 - \alpha)c_0)^2} = \frac{\alpha^2}{4(1 - \alpha)^2} = \frac{0.2198^2}{4(1 - 0.2198)^2} \approx 0.0198$$

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