A homogeneous ideal gaseous reaction $$AB_{2(g)} \rightleftharpoons A_{(g)} + 2B_{(g)}$$ is carried out in a 25 litre flask at 27°C. The initial amount of $$AB_2$$ was 1 mole and the equilibrium pressure was 1.9 atm. The value of $$K_p$$ is $$x \times 10^{-2}$$. The value of $$x$$ is ______ (Integer answer) [R = 0.08206 dm$$^3$$ atm K$$^{-1}$$ mol$$^{-1}$$]

The reaction is $$AB_{2(g)} \rightleftharpoons A_{(g)} + 2B_{(g)}$$.

Initially we have 1 mole of $$AB_2$$ in a 25 L flask at 27°C (300 K). Let $$\alpha$$ be the degree of dissociation.

At equilibrium: moles of $$AB_2 = 1 - \alpha$$, moles of $$A = \alpha$$, moles of $$B = 2\alpha$$. Total moles $$= 1 - \alpha + \alpha + 2\alpha = 1 + 2\alpha$$.

Using the ideal gas law with the equilibrium pressure: $$P_{total} = \frac{n_{total}RT}{V}$$

$$1.9 = \frac{(1 + 2\alpha)(0.08206)(300)}{25}$$

$$1.9 = \frac{(1 + 2\alpha)(24.618)}{25}$$

$$1.9 \times 25 = (1 + 2\alpha)(24.618)$$

$$47.5 = 24.618(1 + 2\alpha)$$

$$1 + 2\alpha = \frac{47.5}{24.618} = 1.9295$$

$$2\alpha = 0.9295$$

$$\alpha = 0.4648$$

Now we find the partial pressures. The total pressure is 1.9 atm and total moles $$= 1 + 2(0.4648) = 1.9295$$.

$$P_{AB_2} = \frac{1 - \alpha}{1 + 2\alpha} \times P_{total} = \frac{0.5352}{1.9295} \times 1.9 = 0.5270 \text{ atm}$$

$$P_A = \frac{\alpha}{1 + 2\alpha} \times P_{total} = \frac{0.4648}{1.9295} \times 1.9 = 0.4576 \text{ atm}$$

$$P_B = \frac{2\alpha}{1 + 2\alpha} \times P_{total} = \frac{0.9295}{1.9295} \times 1.9 = 0.9153 \text{ atm}$$

$$K_p = \frac{P_A \times P_B^2}{P_{AB_2}} = \frac{(0.4576)(0.9153)^2}{0.5270}$$

$$K_p = \frac{(0.4576)(0.8378)}{0.5270} = \frac{0.3834}{0.5270} = 0.7275$$

So $$K_p = 0.7275 = 72.75 \times 10^{-2}$$, giving $$x \approx 73$$.

The answer is $$\boxed{73}$$.