Question

Consider the following isotherms at equilibrium for two disperse dyes \( D_1 \) and \( D_2 \) dyed on polyester. If the partition coefficients of these are \( K_1 \) and \( K_2 \), respectively, the value of \( \frac{K_2}{K_1} \text{ is } \underline{\hspace{2cm}}. \)

Hint:

The partition coefficient is the ratio of the concentration of a substance in one phase (e.g., fibre) to its concentration in another phase (e.g., solution).

Answer 1

Correct Answer: 4

From the graph, the partition coefficient \( K \) is given by the ratio of dye concentration on the fibre to the concentration in solution. For dye \( D_1 \), the ratio \( K_1 \) is given by: \[ K_1 = \frac{\text{Dye on fibre}}{\text{Dye in solution}}. \] For dye \( D_2 \), the ratio \( K_2 \) is given by: \[ K_2 = \frac{\text{Dye on fibre}}{\text{Dye in solution}}. \] From the graph: \[ K_1 = \frac{5}{0.05} = 100, K_2 = \frac{10}{0.1} = 100. \] Thus: \[ \frac{K_2}{K_1} = \frac{100}{100} = 1. \] Thus, the value of \( \frac{K_2}{K_1} \) is \( 1 \).