The incorrect match in the following is:

To decide which pair is mismatched, we first recall the fundamental thermodynamic relation that links the standard Gibbs free-energy change $$\Delta G^0$$ with the equilibrium constant $$K$$ of the reaction.

We have the formula

$$\Delta G^0 = -\,R\,T \,\ln K$$

where $$R$$ is the universal gas constant and $$T$$ is the absolute temperature (in kelvin). This equation is valid for any reaction at equilibrium under standard conditions.

Now, we examine how the signs and magnitudes of $$\Delta G^0$$ and $$K$$ are connected through this equation.

Because $$R$$ and $$T$$ are always positive, the sign of $$\Delta G^0$$ is governed solely by the sign of $$-\ln K$$. So we rewrite

$$\Delta G^0 = -RT \ln K \;\;\Longrightarrow\;\; \dfrac{\Delta G^0}{-RT} = \ln K$$

Next we convert the natural logarithm statement into exponential form:

$$\ln K = \dfrac{\Delta G^0}{-RT} \;\;\Longrightarrow\;\; K = e^{\frac{\Delta G^0}{-RT}}$$

From this exponential expression we deduce the following three direct consequences.

• If $$\Delta G^0 < 0$$, then $$\dfrac{\Delta G^0}{-RT}$$ is positive, hence $$e^{\text{positive}} > 1$$, so $$K > 1$$.

• If $$\Delta G^0 > 0$$, then $$\dfrac{\Delta G^0}{-RT}$$ is negative, hence $$e^{\text{negative}} < 1$$, so $$K < 1$$.

• If $$\Delta G^0 = 0$$, then $$\dfrac{\Delta G^0}{-RT} = 0$$, hence $$e^0 = 1$$, so $$K = 1$$.

We now compare these correct relationships with each option provided.

A. $$\Delta G^0 < 0,\; K > 1$$   matches the first consequence, therefore correct.

B. $$\Delta G^0 > 0,\; K < 1$$   matches the second consequence, therefore correct.

C. $$\Delta G^0 = 0,\; K = 1$$   matches the third consequence, therefore correct.

D. $$\Delta G^0 < 0,\; K < 1$$   contradicts the first consequence because when $$\Delta G^0 < 0$$ we must have $$K > 1$$, not $$K < 1$$. Hence this pair is mismatched.

So among all the given options, the only incorrect match is Option D.

Hence, the correct answer is Option 4.