# 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:

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

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 $$K_c = 0.01984.$$

• 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 $p_{\ce{I2}} = p_{\mathrm{H2}} = \frac{\alpha}{2} * p_\mathrm{tot}$ Jan 28, 2021 at 15:48

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