1. Given reaction

$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g), \qquad \Delta H = -57.2\,\text{kJ mol}^{-1}$

Since $\Delta H < 0$, the reaction is exothermic.

Also,

$K_c = 1.7 \times 10^{16}$

This is a very large equilibrium constant, so equilibrium strongly favors products.

2. Check option A

Reaction has:

• Reactant moles of gas $= 2+1=3$
• Product moles of gas $=2$

Increase in pressure shifts equilibrium toward the side with fewer moles of gas.

So equilibrium shifts forward (towards $SO_3$).

✅ Option A is correct.

3. Check option B

Adding an inert gas at constant volume does not change the partial pressures of reacting gases because

$p_i = \frac{n_iRT}{V}$

and $n_i, T, V$ for reacting gases remain unchanged.

Also, equilibrium constant depends only on temperature, not on addition of inert gas.

✅ Option B is correct.

4. Check option C

A very large $K_c$ means products are highly favored at equilibrium, i.e. reaction is thermodynamically favorable.

But catalyst does not change equilibrium position or equilibrium constant. It only speeds up the rate at which equilibrium is attained.

So even if reaction is favorable, a catalyst may still be required to make the reaction proceed fast enough in practice.

Hence the statement

"The equilibrium constant is large suggestive of reaction going to completion and so no catalyst is required"

is incorrect.

❌ Option C is incorrect.

5. Check option D

For an exothermic reaction, increasing temperature shifts equilibrium in the backward direction. Thus equilibrium constant decreases with increase in temperature.

From van't Hoff relation:

$\frac{d(\ln K)}{dT} = \frac{\Delta H}{RT^2}$

Since $\Delta H < 0$, this derivative is negative, so $K$ decreases as $T$ increases.

✅ Option D is correct.

6. Final conclusion

The incorrect statement is:

$\boxed{\text{C}}$

This matches the stored correct answer.