At a temperature of $450\,^\circ\text{C}$, the reversible reaction below has reached equilibrium. $\text{H}_2 + \text{I}_2 \rightleftharpoons \text{2HI} \qquad \Delta H = -9.5\,\text{kJ mol}^{-1}$ The table gives the equilibrium concentrations in the reaction mixture. $[\text{H}_2] = 1.02\,\text{mol dm}^{-3}$, $[\text{I}_2] = 0.02\,\text{mol dm}^{-3}$, $[\text{HI}] = 0.98\,\text{mol dm}^{-3}$. The temperature of the reaction mixture is raised while the volume stays constant. Under these conditions, the concentration of one component decreases to $0.80\,\text{mol dm}^{-3}$ at equilibrium. What is the concentration of $\text{H}_2$ in the new equilibrium mixture?
- A$0.80\,\text{mol dm}^{-3}$
- B$0.93\,\text{mol dm}^{-3}$
- C$1.11\,\text{mol dm}^{-3}$
- D$1.20\,\text{mol dm}^{-3}$