One mole of pure hydrogen iodide gas, HI(g), is sealed in a 1 dm³ container at 721 K - NSC Physical Sciences - Question 6 - 2023 - Paper 2

Given the balanced equation, the mole ratio of HI to H2 is 2:1. Initially, we have 1 mole of HI.

• Change in moles of I2: 0.11 moles

• From the balanced equation, for every 2 moles of HI that react, 1 mole of H2 is produced. This means:

  • Moles of HI that reacted = 2 * 0.11 = 0.22 moles

• Therefore, the equilibrium moles of H2 is:

  $0.22 / 2 = 0.11 ext{ moles of H2 at equilibrium.}$

Since 0.22 moles of HI reacted, the remaining moles of HI are:

$1 ext{ mole (initial)} - 0.22 ext{ moles (reacted)} = 0.78 ext{ moles of HI at equilibrium.}$

The forward reaction is ENDOTHERMIC because the equilibrium constant increases as the temperature increases (from 0.02 at 721 K to 0.09 at 850 K). This indicates that the formation of products (H2 and I2) absorbs heat.

According to Le Chatelier's principle, increasing temperature favors the endothermic reaction. Since the equilibrium constant increases with temperature, it confirms that the forward reaction is endothermic, where heat is absorbed to convert HI into H2 and I2.

From the previous calculations, at 850 K, we have 0.63 moles of HI remaining. To find the mass, we use the molar mass of HI, which is approximately 128 g/mol:

$ext{Mass} = ext{moles} imes ext{molar mass} = 0.63 ext{ mol} imes 128 ext{ g/mol} = 80.64 ext{ g}$

Thus, the mass of HI at the new equilibrium is 80.64 g.