At a certain temperature and total pressure of 105Pa, iodine vapour contains 40% by volume of I atoms , calculate the K

To calculate the equilibrium constant (K) for the dissociation of iodine vapor into its atomic form, we first need to understand the chemical equilibrium involved in this process. The dissociation reaction can be represented as follows:

Dissociation Reaction

The dissociation of iodine (I₂) into iodine atoms (I) can be expressed by the equation:

I₂(g) ⇌ 2 I(g)

Understanding the Given Data

We know that at a total pressure of 105 Pa, the iodine vapor contains 40% by volume of I atoms. This means that 40% of the total pressure is contributed by the iodine atoms, while the remaining 60% is due to the molecular iodine (I₂).

Calculating Partial Pressures

To find the partial pressures of I₂ and I, we can use the total pressure and the volume percentages:

• Partial pressure of I: 40% of 105 Pa = 0.40 × 105 Pa = 42 Pa
• Partial pressure of I₂: 60% of 105 Pa = 0.60 × 105 Pa = 63 Pa

Setting Up the Equilibrium Expression

The equilibrium constant (K) for the reaction can be expressed in terms of the partial pressures of the products and reactants:

K = (P_I)² / (P_I2)

Substituting the values we calculated:

K = (42 Pa)² / (63 Pa)

Performing the Calculation

Now, let's calculate K:

• (42 Pa)² = 1764 Pa²
• K = 1764 Pa² / 63 Pa = 28 Pa

Final Result

Thus, the equilibrium constant K for the dissociation of iodine vapor at the given conditions is:

K = 28 Pa

This value indicates the extent to which iodine dissociates into its atomic form under the specified conditions. A higher K value would suggest a greater degree of dissociation, while a lower K would indicate that the molecular form predominates. Understanding these relationships is crucial in fields such as chemical engineering and atmospheric chemistry, where the behavior of gases is essential.