Question

Using partial pressure, demonstrate that for the gaseous reaction of the type XY(g) = X(s) + Y(s) at a given temperature, the pressure at which XY is exactly one quarter dissociated is numerically equal to fifteen times the equilibrium constant.

Answer 1

The equilibrium constant expression for the gaseous reaction XY(g) = X(s) + Y(s) in terms of partial pressures is Kp = (P_X)(P_Y)/(P_XY). The pressure at which XY is one quarter dissociated, P_quarter, is numerically equal to fifteen times the equilibrium constant K.

Step-by-step explanation:

For the gaseous reaction XY(g) = X(s) + Y(s), the equilibrium constant expression in terms of partial pressures is Kp = (P_X)(P_Y)/(P_XY), where P_X, P_Y, and P_XY are the partial pressures of X, Y, and XY, respectively. If the pressure at which XY is exactly one quarter dissociated is P_quarter, then the equilibrium constant K is numerically equal to P_quarter multiplied by fifteen, so K = 15 * P_quarter.