Write the equilibrium constant expression for C(s) + CO2(g) + 2Cl2(g) <=> 2COCl2(g) and calculate Kc using the given reactions.

Multiply reaction (2) by 2:
$$
2CO(g) + 2Cl_2(g) \rightleftharpoons 2COCl_2(g)
$$
The new equilibrium constant for this reaction is $(K_{c2})^2$.

$$
C(s) + CO_2(g) \rightleftharpoons 2CO(g)
$$
$$
2CO(g) + 2Cl_2(g) \rightleftharpoons 2COCl_2(g)
$$
Summing these equations and canceling $2CO(g)$ from both sides gives:
$$
C(s) + CO_2(g) + 2Cl_2(g) \rightleftharpoons 2COCl_2(g)
$$

When reactions are added, their equilibrium constants are multiplied.
$$
K_c = K_{c1} \times (K_{c2})^2
$$
$$
K_c = (1.4 \times 10^{12}) \times (5.5 \times 10^1)^2
$$
$$
K_c = (1.4 \times 10^{12}) \times (30.25 \times 10^2)
$$
$$
K_c = (1.4 \times 10^{12}) \times (3.025 \times 10^3)
$$
$$
K_c = (1.4 \times 3.025) \times 10^{12+3}
$$
$$
K_c = 4.235 \times 10^{15}
$$
The equilibrium constant for the reaction is $\mathbf{4.235 \times 10^{15}}$.