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(D) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the equation: $\Delta G^{\circ} = -RT \ln K$.

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$	ext{equilibrium constant } (K) = 1$

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(D) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the equation: $\Delta G^{\circ} = -RT \ln K$. At standard states,the reaction quotient $Q$ is equal to $1$. At equilibrium,$\Delta G = 0$. However,the question specifies standard states,implying $\Delta G^{\circ} = -RT \ln K$. For the reaction to be at equilibrium under standard conditions,$\Delta G^{\circ}$ must be $0$. Therefore,$0 = -RT \ln K$,which implies $\ln K = 0$. Thus,$K = e^0 = 1$.

When a reaction is carried out at standard states,then at equilibrium:

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