Q.At a given temperature and pressure, the equilibrium constant values for the equilibria

[!FORMULA] $3A_2 + B_2 + 2C \rightleftharpoons 2A_3BC \quad (K_1)$

[!FORMULA] $A_3BC \rightleftharpoons \frac{3}{2}A_2 + \frac{1}{2}B_2 + C \quad (K_2)$

The relation between $K_1$ and $K_2$ is

When several equilibria share species, their equilibrium constants combine algebraically in step with however the reactions themselves are added, reversed, multiplied or divided.

Reversing a reaction inverts its constant. If the forward reaction $xA+yB\rightleftharpoons lC+mD$ has equilibrium constant $K_C$, the reverse reaction $lC+mD\rightleftharpoons xA+yB$ has constant $K_C'=1/K_C$ (swapping numerator and denominator inverts the ratio). This is why the constant for the dissociation of one mole of $SO_3$ into $SO_2$ and $O_2$ is $(1/K_1)^{1/2}$, not simply $1/K_1$, when $K_1$ is defined for the two-mole formation reaction $2SO_2+O_2\rightleftharpoons2SO_3$: dissociating one mole of $SO_3$ is half of the reverse of the two-mole reaction, so its constant is the square root of the reverse constant, $\sqrt{1/K_1}$.

Adding reactions multiplies their constants. If $A\rightleftharpoons B$ has constant $K_1$ and $B\rightleftharpoons C$ has constant $K_2$, then the sum reaction $A\rightleftharpoons C$ (obtained by adding the two equations and cancelling the common intermediate B) has constant $K=K_1K_2$, because each intermediate species' concentration term cancels out of the product of the two separate expressions. Chained over three or more steps, this becomes $K_4=K_1K_2K_3$ for a four-species chain $A\rightleftharpoons B\rightleftharpoons C\rightleftharpoons D$ compared against the single direct step $A\rightleftharpoons D$.

Scaling a reaction by n raises its constant to the n-th power. If $A\rightleftharpoons B$ has constant $K$, then $2A\rightleftharpoons 2B$ (the same reaction written with every coefficient doubled) has constant $K^2$, since every concentration term in the expression is itself squared.

Putting reversal, addition and scaling together lets a target reaction's constant be built up from other reactions' known constants without ever needing new experimental data -- for instance, given $K_1$ for $N_2+3H_2\rightleftharpoons2NH_3$, $K_2$ for $N_2+O_2\rightleftharpoons2NO$ and $K_3$ for $H_2+\tfrac12O_2\rightleftharpoons H_2O$, the constant for $2NH_3+\tfrac52O_2\rightleftharpoons2NO+3H_2O$ is found by reversing reaction 1 (giving $1/K_1$), keeping reaction 2 as is (giving $K_2$), and tripling reaction 3 (giving $K_3^3$), then multiplying: $K=\dfrac{K_2K_3^3}{K_1}$.