For the following reaction, equilibrium constants are given:
$$S(s) + O_2(g) \rightleftharpoons SO_2(g); \quad K_1 = 10^{52}$$
$$2S(s) + 3O_2(g) \rightleftharpoons 2SO_3(g); \quad K_2 = 10^{129}$$
The equilibrium constant for the reaction, $$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$$ is:

We have two given equilibria:

$$S(s) + O_2(g) \rightleftharpoons SO_2(g), \qquad K_1 = 10^{52}$$

$$2S(s) + 3O_2(g) \rightleftharpoons 2SO_3(g), \qquad K_2 = 10^{129}$$

Our task is to find the equilibrium constant for

$$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$$

First, we recall two rules for equilibrium constants:

1. If an equation is reversed, its new equilibrium constant is the reciprocal of the original, $$K_{\text{new}} = \dfrac{1}{K_{\text{old}}}\,.$$

2. If two (or more) equations are added, the equilibrium constant for the overall reaction equals the product of the individual constants, $$K_{\text{overall}} = K_1 \times K_2 \times \dots$$

Now we manipulate the given reactions so that, when added, they produce the desired equation.

Step 1 Double the first equilibrium to get exactly two moles of sulphur dioxide. Doubling a reaction squares its equilibrium constant:

$$2\bigl[S(s) + O_2(g) \rightleftharpoons SO_2(g)\bigr]$$

gives

$$2S(s) + 2O_2(g) \rightleftharpoons 2SO_2(g), \qquad K_1^{\,2} = \left(10^{52}\right)^2 = 10^{104}.$$

Step 2 Reverse this doubled equation, because in our target reaction $$2SO_2(g)$$ must appear on the left. Reversing gives the reciprocal of the constant:

$$2SO_2(g) \rightleftharpoons 2S(s) + 2O_2(g), \qquad K_{\text{rev}} = \dfrac{1}{10^{104}} = 10^{-104}.$$

Step 3 Keep the second given equilibrium unchanged:

$$2S(s) + 3O_2(g) \rightleftharpoons 2SO_3(g), \qquad K_2 = 10^{129}.$$

Step 4 Add the reversed, doubled first equation to the second equation. Adding the left and right sides we obtain

$$\bigl[2SO_2(g)\bigr] + \bigl[2S(s) + 3O_2(g)\bigr] \;\longrightarrow\; \bigl[2S(s) + 2O_2(g)\bigr] + \bigl[2SO_3(g)\bigr].$$

The solid sulphur $$2S(s)$$ appears on both sides and cancels. Subtracting $$2O_2(g)$$ from each side leaves

$$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g),$$

exactly the required reaction.

According to the multiplication rule, the equilibrium constant for the overall reaction is

$$K_3 = K_{\text{rev}} \times K_2 = 10^{-104} \times 10^{129} = 10^{25}.$$

Hence, the correct answer is Option B.