Hydrogen and iodine are able to react reversibly to form hydrogen iodide. The equation is given below. $\text{H}_2(\text{g}) + \text{I}_2(\text{g}) \rightleftharpoons 2\text{HI}(\text{g})$ $4.00\,\text{mol}$ of hydrogen gas and $X\,\text{mol}$ of iodine vapour are placed together in a sealed container with a volume of $1.00\,\text{dm}^3$ at $460\,\text{K}$. The mixture is then left until equilibrium is reached. At equilibrium, the mixture contains $2.00\,\text{mol}$ of hydrogen iodide. For this reaction at $460\,\text{K}$, the equilibrium constant, $K_c$, is $4.0$. Determine the value of $X$.
- A$0.50\,\text{mol}$
- B$1.17\,\text{mol}$
- C$1.33\,\text{mol}$
- D$2.50\,\text{mol}$